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CHAPTER
1
First-Order ODEs
Chapter 1 begins the study of ordinary differential equations (ODEs) by deriving them from
physical or other problems (modeling), solving them by standard mathematical methods,
and interpreting solutions and their graphs in terms of a given problem. The simplest ODEs
to be discussed are ODEs of the first order because they involve only the first derivative
of the unknown function and no higher derivatives. These unknown functions will usually
be denoted by y1x2 or y1t2 when the independent variable denotes time t. The chapter ends
with a study of the existence and uniqueness of solutions of ODEs in Sec. 1.7.
Understanding the basics of ODEs requires solving problems by hand (paper and pencil,
or typing on your computer, but first without the aid of a CAS). In doing so, you will
gain an important conceptual understanding and feel for the basic terms, such as ODEs,
direction field, and initial value problem. If you wish, you can use your Computer Algebra
System (CAS) for checking solutions.
COMMENT. Numerics for first-order ODEs can be studied immediately after this
chapter. See Secs. 21.1–21.2, which are independent of other sections on numerics.
Prerequisite: Integral calculus.
Sections that may be omitted in a shorter course: 1.6, 1.7.
References and Answers to Problems: App. 1 Part A, and App. 2.
1.1 Basic Concepts. Modeling
Physical
System
Mathematical
Model
Mathematical
Solution
Physical
Interpretation
Fig. 1. Modeling,
solving, interpreting
2
If we want to solve an engineering problem (usually of a physical nature), we first
have to formulate the problem as a mathematical expression in terms of variables,
functions, and equations. Such an expression is known as a mathematical model of the
given problem. The process of setting up a model, solving it mathematically, and
interpreting the result in physical or other terms is called mathematical modeling or,
briefly, modeling.
Modeling needs experience, which we shall gain by discussing various examples and
problems. (Your computer may often help you in solving but rarely in setting up models.)
Now many physical concepts, such as velocity and acceleration, are derivatives. Hence
a model is very often an equation containing derivatives of an unknown function. Such
a model is called a differential equation. Of course, we then want to find a solution (a
function that satisfies the equation), explore its properties, graph it, find values of it, and
interpret it in physical terms so that we can understand the behavior of the physical system
in our given problem. However, before we can turn to methods of solution, we must first
define some basic concepts needed throughout this chapter.
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SEC. 1.1 Basic Concepts. Modeling
3
y
h
Velocity
v
Water level h
Falling stone
Parachutist
y″ = g = const.
(Sec. 1.1)
mv′ = mg – bv
(Sec. 1.2)
Outflowing water
h′ = –k h
(Sec. 1.3)
2
y
R
(k)
C
E
t
L
y
m
Displacement y
Vibrating mass
on a spring
my″ + ky = 0
(Secs. 2.4, 2.8)
Beats of a vibrating
system
2
y″ + ω
w0 y = cos ω
wt, ω
w0 ≈ ω
w
(Sec. 2.8)
Current I in an
RLC circuit
LI″ + RI′ + 1 I = E′
C
(Sec. 2.9)
L
θ
y
Lotka–Volterra
predator–prey model
Pendulum
y′1 = ay1 – by1 y2
EIy = f(x)
Lθ″ + g sin θθ = 0
y′2 = ky1 y2 – ly2
(Sec. 3.3)
(Sec. 4.5)
(Sec. 4.5)
Deformation of a beam
iv
Fig. 2.
Some applications of differential equations
An ordinary differential equation (ODE) is an equation that contains one or several
derivatives of an unknown function, which we usually call y(x) (or sometimes y(t) if the
independent variable is time t). The equation may also contain y itself, known functions
of x (or t), and constants. For example,
(1)
y r ⫽ cos x
(2)
y s ⫹ 9y ⫽ eⴚ2x
(3)
y r y t ⫺ 32 y r 2 ⫽ 0
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CHAP. 1 First-Order ODEs
are ordinary differential equations (ODEs). Here, as in calculus, y r denotes dy>dx,
y s ⫽ d 2y>dx 2, etc. The term ordinary distinguishes them from partial differential
equations (PDEs), which involve partial derivatives of an unknown function of two
or more variables. For instance, a PDE with unknown function u of two variables x
and y is
0 2u
0x
2
⫹
0 2u
0y 2
⫽ 0.
PDEs have important engineering applications, but they are more complicated than ODEs;
they will be considered in Chap. 12.
An ODE is said to be of order n if the nth derivative of the unknown function y is the
highest derivative of y in the equation. The concept of order gives a useful classification
into ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of second
order, and (3) of third order.
In this chapter we shall consider first-order ODEs. Such equations contain only the
first derivative y r and may contain y and any given functions of x. Hence we can write
them as
(4)
F(x, y, y r ) ⫽ 0
or often in the form
y r ⫽ f (x, y).
This is called the explicit form, in contrast to the implicit form (4). For instance, the implicit
ODE x ⴚ3y r ⫺ 4y 2 ⫽ 0 (where x ⫽ 0) can be written explicitly as y r ⫽ 4x 3y 2.
Concept of Solution
A function
y ⫽ h(x)
is called a solution of a given ODE (4) on some open interval a ⬍ x ⬍ b if h(x) is
defined and differentiable throughout the interval and is such that the equation becomes
an identity if y and y r are replaced with h and h r , respectively. The curve (the graph) of
h is called a solution curve.
Here, open interval a ⬍ x ⬍ b means that the endpoints a and b are not regarded as
points belonging to the interval. Also, a ⬍ x ⬍ b includes infinite intervals ⫺⬁ ⬍ x ⬍ b,
a ⬍ x ⬍ ⬁, ⫺⬁ ⬍ x ⬍ ⬁ (the real line) as special cases.
EXAMPLE 1
Verification of Solution
Verify that y ⫽ c>x (c an arbitrary constant) is a solution of the ODE xy r ⫽ ⫺y for all x ⫽ 0. Indeed, differentiate
y ⫽ c>x to get y r ⫽ ⫺c>x 2. Multiply this by x, obtaining xy r ⫽ ⫺c>x; thus, xy r ⫽ ⫺y, the given ODE.
䊏
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SEC. 1.1 Basic Concepts. Modeling
EXAMPLE 2
5
Solution by Calculus. Solution Curves
The ODE y r ⫽ dy>dx ⫽ cos x can be solved directly by integration on both sides. Indeed, using calculus,
we obtain y ⫽ 兰 cos x dx ⫽ sin x ⫹ c, where c is an arbitrary constant. This is a family of solutions. Each value
of c, for instance, 2.75 or 0 or ⫺8, gives one of these curves. Figure 3 shows some of them, for c ⫽ ⫺3, ⫺2,
⫺1, 0, 1, 2, 3, 4.
䊏
y
4
2
–π
π
0
x
2π
–2
–4
Fig. 3.
EXAMPLE 3
Solutions y ⫽ sin x ⫹ c of the ODE y r ⫽ cos x
(A) Exponential Growth. (B) Exponential Decay
From calculus we know that y ⫽ ce0.2t has the derivative
yr ⫽
dy
dt
⫽ 0.2e0.2t ⫽ 0.2y.
Hence y is a solution of y r ⫽ 0.2y (Fig. 4A). This ODE is of the form y r ⫽ ky. With positive-constant k it can
model exponential growth, for instance, of colonies of bacteria or populations of animals. It also applies to
humans for small populations in a large country (e.g., the United States in early times) and is then known as
Malthus’s law.1 We shall say more about this topic in Sec. 1.5.
(B) Similarly, y r ⫽ ⫺0.2 (with a minus on the right) has the solution y ⫽ ceⴚ0.2t, (Fig. 4B) modeling
exponential decay, as, for instance, of a radioactive substance (see Example 5).
䊏
y
y
40
2.5
2.0
30
1.5
20
1.0
10
0
0.5
0
2
4
6
8
10
12
14 t
Fig. 4A. Solutions of y r ⫽ 0.2y
in Example 3 (exponential growth)
1
0
0
2
4
6
8
10
12
14 t
Fig. 4B. Solutions of y r ⫽ ⫺0.2y
in Example 3 (exponential decay)
Named after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766–1834).
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CHAP. 1 First-Order ODEs
We see that each ODE in these examples has a solution that contains an arbitrary
constant c. Such a solution containing an arbitrary constant c is called a general solution
of the ODE.
(We shall see that c is sometimes not completely arbitrary but must be restricted to some
interval to avoid complex expressions in the solution.)
We shall develop methods that will give general solutions uniquely (perhaps except for
notation). Hence we shall say the general solution of a given ODE (instead of a general
solution).
Geometrically, the general solution of an ODE is a family of infinitely many solution
curves, one for each value of the constant c. If we choose a specific c (e.g., c ⫽ 6.45 or 0
or ⫺2.01) we obtain what is called a particular solution of the ODE. A particular solution
does not contain any arbitrary constants.
In most cases, general solutions exist, and every solution not containing an arbitrary
constant is obtained as a particular solution by assigning a suitable value to c. Exceptions
to these rules occur but are of minor interest in applications; see Prob. 16 in Problem
Set 1.1.
Initial Value Problem
In most cases the unique solution of a given problem, hence a particular solution, is
obtained from a general solution by an initial condition y(x 0) ⫽ y0, with given values
x 0 and y0, that is used to determine a value of the arbitrary constant c. Geometrically
this condition means that the solution curve should pass through the point (x 0, y0)
in the xy-plane. An ODE, together with an initial condition, is called an initial value
problem. Thus, if the ODE is explicit, y r ⫽ f (x, y), the initial value problem is of
the form
(5)
EXAMPLE 4
y r ⫽ f (x, y),
y(x 0) ⫽ y0.
Initial Value Problem
Solve the initial value problem
yr ⫽
dy
dx
⫽ 3y,
y(0) ⫽ 5.7.
The general solution is y(x) ⫽ ce3x; see Example 3. From this solution and the initial condition
we obtain y(0) ⫽ ce0 ⫽ c ⫽ 5.7. Hence the initial value problem has the solution y(x) ⫽ 5.7e3x. This is a
particular solution.
䊏
Solution.
More on Modeling
The general importance of modeling to the engineer and physicist was emphasized at the
beginning of this section. We shall now consider a basic physical problem that will show
the details of the typical steps of modeling. Step 1: the transition from the physical situation
(the physical system) to its mathematical formulation (its mathematical model); Step 2:
the solution by a mathematical method; and Step 3: the physical interpretation of the result.
This may be the easiest way to obtain a first idea of the nature and purpose of differential
equations and their applications. Realize at the outset that your computer (your CAS)
may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work.
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SEC. 1.1 Basic Concepts. Modeling
7
And Step 2 requires a solid knowledge and good understanding of solution methods
available to you—you have to choose the method for your work by hand or by the
computer. Keep this in mind, and always check computer results for errors (which may
arise, for instance, from false inputs).
EXAMPLE 5
Radioactivity. Exponential Decay
Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.
Physical Information. Experiments show that at each instant a radioactive substance decomposes—and is thus
decaying in time—proportional to the amount of substance present.
Step 1. Setting up a mathematical model of the physical process. Denote by y(t) the amount of substance still
present at any time t. By the physical law, the time rate of change y r (t) ⫽ dy>dt is proportional to y(t). This
gives the first-order ODE
dy
(6)
dt
⫽ ⫺ky
where the constant k is positive, so that, because of the minus, we do get decay (as in [B] of Example 3).
The value of k is known from experiments for various radioactive substances (e.g., k ⫽ 1.4 ⴢ 10ⴚ11 sec ⴚ1,
approximately, for radium 226
88 Ra).
Now the given initial amount is 0.5 g, and we can call the corresponding instant t ⫽ 0. Then we have the
initial condition y(0) ⫽ 0.5. This is the instant at which our observation of the process begins. It motivates
the term initial condition (which, however, is also used when the independent variable is not time or when
we choose a t other than t ⫽ 0). Hence the mathematical model of the physical process is the initial value
problem
dy
(7)
dt
⫽ ⫺ky,
y(0) ⫽ 0.5.
Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decay
and has the general solution (with arbitrary constant c but definite given k)
y(t) ⫽ ceⴚkt.
(8)
We now determine c by using the initial condition. Since y(0) ⫽ c from (8), this gives y(0) ⫽ c ⫽ 0.5. Hence
the particular solution governing our process is (cf. Fig. 5)
y(t) ⫽ 0.5eⴚkt
(9)
(k ⬎ 0).
Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!)
that your solution (9) satisfies (7) as well as y(0) ⫽ 0.5:
dy
dt
⫽ ⫺0.5keⴚkt ⫽ ⫺k ⴢ 0.5eⴚkt ⫽ ⫺ky,
y(0) ⫽ 0.5e0 ⫽ 0.5.
Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts from
䊏
the correct initial amount and decreases with time because k is positive. The limit of y as t : ⬁ is zero.
y
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
Fig. 5. Radioactivity (Exponential decay,
y ⫽ 0.5e⫺kt, with k ⫽ 1.5 as an example)
t
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8
CHAP. 1 First-Order ODEs
PROBLEM SET 1.1
1–8
CALCULUS
17–20
Solve the ODE by integration or by remembering a
differentiation formula.
1. y r ⫹ 2 sin 2 px ⫽ 0
2
2. y r ⫹ xeⴚx >2 ⫽ 0
3. y r ⫽ y
4. y r ⫽ ⫺1.5y
5. y r ⫽ 4eⴚx cos x
6. y s ⫽ ⫺y
7. y r ⫽ cosh 5.13x
8. y t ⫽ eⴚ0.2x
9–15
VERIFICATION. INITIAL VALUE
PROBLEM (IVP)
(a) Verify that y is a solution of the ODE. (b) Determine
from y the particular solution of the IVP. (c) Graph the
solution of the IVP.
9. y r ⫹ 4y ⫽ 1.4, y ⫽ ceⴚ4x ⫹ 0.35, y(0) ⫽ 2
2
10. y r ⫹ 5xy ⫽ 0, y ⫽ ceⴚ2.5x , y(0) ⫽ p
11. y r ⫽ y ⫹ ex, y ⫽ (x ⫹ c)ex, y(0) ⫽ 12
12. yy r ⫽ 4x, y 2 ⫺ 4x 2 ⫽ c (y ⬎ 0), y(1) ⫽ 4
1
13. y r ⫽ y ⫺ y 2, y ⫽
, y(0) ⫽ 0.25
1 ⫹ ceⴚx
14. y r tan x ⫽ 2y ⫺ 8, y ⫽ c sin 2 x ⫹ 4, y(12 p) ⫽ 0
15. Find two constant solutions of the ODE in Prob. 13 by
inspection.
16. Singular solution. An ODE may sometimes have an
additional solution that cannot be obtained from the
general solution and is then called a singular solution.
The ODE y r 2 ⫺ xy r ⫹ y ⫽ 0 is of this kind. Show
by differentiation and substitution that it has the
general solution y ⫽ cx ⫺ c2 and the singular solution
y ⫽ x 2>4. Explain Fig. 6.
y
3
2
1
–4
Fig. 6.
–2 –1
–2
–3
–4
–5
2
4 x
Particular solutions and singular
solution in Problem 16
MODELING, APPLICATIONS
These problems will give you a first impression of modeling.
Many more problems on modeling follow throughout this
chapter.
17. Half-life. The half-life measures exponential decay.
It is the time in which half of the given amount of
radioactive substance will disappear. What is the halflife of 226
88 Ra (in years) in Example 5?
18. Half-life. Radium 224
88 Ra has a half-life of about
3.6 days.
(a) Given 1 gram, how much will still be present after
1 day?
(b) After 1 year?
19. Free fall. In dropping a stone or an iron ball, air
resistance is practically negligible. Experiments
show that the acceleration of the motion is constant
(equal to g ⫽ 9.80 m>sec2 ⫽ 32 ft>sec 2, called the
acceleration of gravity). Model this as an ODE for
y(t), the distance fallen as a function of time t. If the
motion starts at time t ⫽ 0 from rest (i.e., with velocity
v ⫽ y r ⫽ 0), show that you obtain the familiar law of
free fall
y ⫽ 12 gt 2.
20. Exponential decay. Subsonic flight. The efficiency
of the engines of subsonic airplanes depends on air
pressure and is usually maximum near 35,000 ft.
Find the air pressure y(x) at this height. Physical
information. The rate of change y r (x) is proportional
to the pressure. At 18,000 ft it is half its value
y0 ⫽ y(0) at sea level. Hint. Remember from calculus
that if y ⫽ ekx, then y r ⫽ kekx ⫽ ky. Can you see
without calculation that the answer should be close
to y0>4?
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SEC. 1.2 Geometric Meaning of y⬘ ⫽ ƒ(x, y). Direction Fields, Euler’s Method
9
1.2 Geometric Meaning of y r ⫽ f (x, y).
Direction Fields, Euler’s Method
A first-order ODE
y r ⫽ f (x, y)
(1)
has a simple geometric interpretation. From calculus you know that the derivative y r (x) of
y(x) is the slope of y(x). Hence a solution curve of (1) that passes through a point (x 0, y0)
must have, at that point, the slope y r (x 0) equal to the value of f at that point; that is,
y r (x 0) ⫽ f (x 0, y0).
Using this fact, we can develop graphic or numeric methods for obtaining approximate
solutions of ODEs (1). This will lead to a better conceptual understanding of an ODE (1).
Moreover, such methods are of practical importance since many ODEs have complicated
solution formulas or no solution formulas at all, whereby numeric methods are needed.
Graphic Method of Direction Fields. Practical Example Illustrated in Fig. 7. We
can show directions of solution curves of a given ODE (1) by drawing short straight-line
segments (lineal elements) in the xy-plane. This gives a direction field (or slope field)
into which you can then fit (approximate) solution curves. This may reveal typical
properties of the whole family of solutions.
Figure 7 shows a direction field for the ODE
yr ⫽ y ⫹ x
(2)
obtained by a CAS (Computer Algebra System) and some approximate solution curves
fitted in.
y
2
1
–2
–1.5
–1
–0.5
0.5
1 x
–1
–2
Fig. 7.
Direction field of y r ⫽ y ⫹ x, with three approximate solution
curves passing through (0, 1), (0, 0), (0, ⫺1), respectively
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CHAP. 1 First-Order ODEs
If you have no CAS, first draw a few level curves f (x, y) ⫽ const of f (x, y), then parallel
lineal elements along each such curve (which is also called an isocline, meaning a curve
of equal inclination), and finally draw approximation curves fit to the lineal elements.
We shall now illustrate how numeric methods work by applying the simplest numeric
method, that is Euler’s method, to an initial value problem involving ODE (2). First we
give a brief description of Euler’s method.
Numeric Method by Euler
Given an ODE (1) and an initial value y(x 0) ⫽ y0, Euler’s method yields approximate
solution values at equidistant x-values x 0, x 1 ⫽ x 0 ⫹ h, x 2 ⫽ x 0 ⫹ 2h, Á , namely,
y1 ⫽ y0 ⫹ hf (x 0, y0)
(Fig. 8)
y2 ⫽ y1 ⫹ hf (x 1, y1),
etc.
In general,
yn ⫽ yn⫺1 ⫹ hf (x n⫺1, yn⫺1)
where the step h equals, e.g., 0.1 or 0.2 (as in Table 1.1) or a smaller value for greater
accuracy.
y
Solution curve
y(x1)
Error of y1
y1
hf(x0, y0)
y0
h
x0
Fig. 8.
x1
x
First Euler step, showing a solution curve, its tangent at (x0, y0),
step h and increment hf (x0, y0) in the formula for y1
Table 1.1 shows the computation of n ⫽ 5 steps with step h ⫽ 0.2 for the ODE (2) and
initial condition y(0) ⫽ 0, corresponding to the middle curve in the direction field. We
shall solve the ODE exactly in Sec. 1.5. For the time being, verify that the initial value
problem has the solution y ⫽ ex ⫺ x ⫺ 1. The solution curve and the values in Table 1.1
are shown in Fig. 9. These values are rather inaccurate. The errors y(x n) ⫺ yn are shown
in Table 1.1 as well as in Fig. 9. Decreasing h would improve the values, but would soon
require an impractical amount of computation. Much better methods of a similar nature
will be discussed in Sec. 21.1.
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SEC. 1.2 Geometric Meaning of y⬘ ⫽ ƒ(x, y). Direction Fields, Euler’s Method
11
Table 1.1. Euler method for y r ⴝ y ⴙ x, y (0) ⴝ 0 for
x ⴝ 0, Á , 1.0 with step h ⴝ 0.2
n
xn
yn
y(x n)
Error
0
1
2
3
4
5
0.0
0.2
0.4
0.6
0.8
1.0
0.000
0.000
0.04
0.128
0.274
0.488
0.000
0.021
0.092
0.222
0.426
0.718
0.000
0.021
0.052
0.094
0.152
0.230
y
0.7
0.5
0.3
0.1
0
Fig. 9.
0.2
0.4
0.6
0.8
1
x
Euler method: Approximate values in Table 1.1 and solution curve
PROBLEM SET 1.2
1–8
DIRECTION FIELDS, SOLUTION CURVES
Graph a direction field (by a CAS or by hand). In the field
graph several solution curves by hand, particularly those
passing through the given points (x, y).
1. y r ⫽ 1 ⫹ y 2, (14 p, 1)
2. yy r ⫹ 4x ⫽ 0, (1, 1), (0, 2)
3. y r ⫽ 1 ⫺ y 2, (0, 0), (2, 12 )
4. y r ⫽ 2y ⫺ y 2, (0, 0), (0, 1), (0, 2), (0, 3)
5. y r ⫽ x ⫺ 1>y, (1, 12 )
6. y r ⫽ sin 2 y, (0, ⫺0.4), (0, 1)
7. y r ⫽ ey>x, (2, 2), (3, 3)
8. y r ⫽ ⫺2xy, (0, 12 ), (0, 1), (0, 2)
of equal inclination) of an autonomous ODE look like?
Give reason.
12–15
MOTIONS
Model the motion of a body B on a straight line with
velocity as given, y(t) being the distance of B from a point
y ⫽ 0 at time t. Graph a direction field of the model (the
ODE). In the field sketch the solution curve satisfying the
given initial condition.
12. Product of velocity times distance constant, equal to 2,
y(0) ⫽ 2.
13. Distance ⫽ Velocity ⫻ Time,
y(1) ⫽ 1
ACCURACY OF DIRECTION FIELDS
14. Square of the distance plus square of the velocity equal
to 1, initial distance 1> 12
Direction fields are very useful because they can give you
an impression of all solutions without solving the ODE,
which may be difficult or even impossible. To get a feel for
the accuracy of the method, graph a field, sketch solution
curves in it, and compare them with the exact solutions.
9. y r ⫽ cos px
10. y r ⫽ ⫺5y 1>2 (Sol. 1y ⫹ 52 x ⫽ c)
11. Autonomous ODE. This means an ODE not showing
x (the independent variable) explicitly. (The ODEs in
Probs. 6 and 10 are autonomous.) What will the level
curves f (x, y) ⫽ const (also called isoclines ⫽ curves
15. Parachutist. Two forces act on a parachutist, the
attraction by the earth mg (m ⫽ mass of person plus
equipment, g ⫽ 9.8 m>sec2 the acceleration of gravity)
and the air resistance, assumed to be proportional to the
square of the velocity v(t). Using Newton’s second law
of motion (mass ⫻ acceleration ⫽ resultant of the forces),
set up a model (an ODE for v(t)). Graph a direction field
(choosing m and the constant of proportionality equal to 1).
Assume that the parachute opens when v ⫽ 10 m>sec.
Graph the corresponding solution in the field. What is the
limiting velocity? Would the parachute still be sufficient
if the air resistance were only proportional to v(t)?
9–10
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CHAP. 1 First-Order ODEs
16. CAS PROJECT. Direction Fields. Discuss direction
fields as follows.
(a) Graph portions of the direction field of the ODE (2)
(see Fig. 7), for instance, ⫺5 ⬉ x ⬉ 2, ⫺1 ⬉ y ⬉ 5.
Explain what you have gained by this enlargement of
the portion of the field.
(b) Using implicit differentiation, find an ODE with
the general solution x 2 ⫹ 9y 2 ⫽ c (y ⬎ 0). Graph its
direction field. Does the field give the impression
that the solution curves may be semi-ellipses? Can you
do similar work for circles? Hyperbolas? Parabolas?
Other curves?
(c) Make a conjecture about the solutions of y r ⫽ ⫺x>y
from the direction field.
(d) Graph the direction field of y r ⫽ ⫺12 y and some
solutions of your choice. How do they behave? Why
do they decrease for y ⬎ 0?
17–20
EULER’S METHOD
This is the simplest method to explain numerically solving
an ODE, more precisely, an initial value problem (IVP).
(More accurate methods based on the same principle are
explained in Sec. 21.1.) Using the method, to get a feel for
numerics as well as for the nature of IVPs, solve the IVP
numerically with a PC or a calculator, 10 steps. Graph the
computed values and the solution curve on the same
coordinate axes.
17. y r ⫽ y,
y(0) ⫽ 1,
h ⫽ 0.1
18. y r ⫽ y,
y(0) ⫽ 1,
h ⫽ 0.01
19. y r ⫽ (y ⫺ x) , y(0) ⫽ 0,
Sol. y ⫽ x ⫺ tanh x
h ⫽ 0.1
20. y r ⫽ ⫺5x 4y 2, y(0) ⫽ 1,
Sol. y ⫽ 1>(1 ⫹ x)5
h ⫽ 0.2
2
1.3 Separable ODEs. Modeling
Many practically useful ODEs can be reduced to the form
g(y) y r ⫽ f (x)
(1)
by purely algebraic manipulations. Then we can integrate on both sides with respect to x,
obtaining
冮 g(y) yrdx ⫽ 冮 f (x) dx ⫹ c.
(2)
On the left we can switch to y as the variable of integration. By calculus, y r dx ⫽ dy, so that
冮 g(y) dy ⫽ 冮 f (x) dx ⫹ c.
(3)
If f and g are continuous functions, the integrals in (3) exist, and by evaluating them we
obtain a general solution of (1). This method of solving ODEs is called the method of
separating variables, and (1) is called a separable equation, because in (3) the variables
are now separated: x appears only on the right and y only on the left.
EXAMPLE 1
Separable ODE
The ODE y r ⫽ 1 ⫹ y 2 is separable because it can be written
dy
1 ⫹ y2
⫽ dx.
By integration,
arctan y ⫽ x ⫹ c
or
y ⫽ tan (x ⫹ c).
It is very important to introduce the constant of integration immediately when the integration is performed.
If we wrote arctan y ⫽ x, then y ⫽ tan x, and then introduced c, we would have obtained y ⫽ tan x ⫹ c, which
䊏
is not a solution (when c ⫽ 0). Verify this.
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SEC. 1.3 Separable ODEs. Modeling
EXAMPLE 2
13
Separable ODE
The ODE y r ⫽ (x ⫹ 1)eⴚxy 2 is separable; we obtain y ⴚ2 dy ⫽ (x ⫹ 1)eⴚx dx.
By integration,
EXAMPLE 3
⫺y ⴚ1 ⫽ ⫺(x ⫹ 2)eⴚx ⫹ c,
y⫽
1
.
(x ⫹ 2)e⫺x ⫺ c
䊏
Initial Value Problem (IVP). Bell-Shaped Curve
Solve y r ⫽ ⫺2xy, y(0) ⫽ 1.8.
Solution.
By separation and integration,
dy
y
⫽ ⫺2x dx,
ln y ⫽ ⫺x 2 ⫹ 苲
c,
y ⫽ ceⴚx .
2
This is the general solution. From it and the initial condition, y(0) ⫽ ce0 ⫽ c ⫽ 1.8. Hence the IVP has the
2
solution y ⫽ 1.8eⴚx . This is a particular solution, representing a bell-shaped curve (Fig. 10).
䊏
y
1
–2
–1
0
1
2 x
Fig. 10. Solution in Example 3 (bell-shaped curve)
Modeling
The importance of modeling was emphasized in Sec. 1.1, and separable equations yield
various useful models. Let us discuss this in terms of some typical examples.
EXAMPLE 4
Radiocarbon Dating2
In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found in
the ice of the Oetztal Alps (hence the name “Oetzi”) in Southern Tyrolia near the Austrian–Italian border, caused
a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon 146 C to carbon 126 C in
this mummy is 52.5% of that of a living organism?
Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon 146 C (made
radioactive by cosmic rays) to ordinary carbon 126 C is constant. When an organism dies, its absorption of 146 C
by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive
carbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of 146 C, which
is 5715 years (CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page 11–52,
line 9).
Modeling. Radioactive decay is governed by the ODE y r ⫽ ky (see Sec. 1.1, Example 5). By
separation and integration (where t is time and y0 is the initial ratio of 146 C to 126 C)
Solution.
dy
y
2
⫽ k dt,
ln ƒ y ƒ ⫽ kt ⫹ c,
y ⫽ y0 ekt
(y0 ⫽ ec).
Method by WILLARD FRANK LIBBY (1908–1980), American chemist, who was awarded for this work
the 1960 Nobel Prize in chemistry.
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CHAP. 1 First-Order ODEs
Next we use the half-life H ⫽ 5715 to determine k. When t ⫽ H, half of the original substance is still present. Thus,
y0ekH ⫽ 0.5y0,
ekH ⫽ 0.5,
k⫽
0.693
ln 0.5
⫽⫺
⫽ ⫺0.0001 213.
H
5715
Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),
ekt ⫽ eⴚ0.0001 213t ⫽ 0.525,
t⫽
ln 0.525
⫽ 5312.
⫺0.0001 213
Answer:
About 5300 years ago.
Other methods show that radiocarbon dating values are usually too small. According to recent research, this is
due to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing. 䊏
EXAMPLE 5
Mixing Problem
Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model
involving a single tank. The tank in Fig. 11 contains 1000 gal of water in which initially 100 lb of salt is dissolved.
Brine runs in at a rate of 10 gal> min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank is
kept uniform by stirring. Brine runs out at 10 gal> min. Find the amount of salt in the tank at any time t.
Solution.
Step 1. Setting up a model. Let y(t) denote the amount of salt in the tank at time t. Its time rate
of change is
y r ⫽ Salt inflow rate ⫺ Salt outflow rate
Balance law.
5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This is 10>1000 ⫽ 0.01
(⫽ 1%) of the total brine content in the tank, hence 0.01 of the salt content y(t), that is, 0.01 y(t). Thus the
model is the ODE
y r ⫽ 50 ⫺ 0.01y ⫽ ⫺0.01(y ⫺ 5000).
(4)
Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on both
sides gives
dy
y ⫺ 5000
⫽ ⫺0.01 dt,
y ⫺ 5000 ⫽ ceⴚ0.01t.
ln ƒ y ⫺ 5000 ƒ ⫽ ⫺0.01t ⫹ c*,
Initially the tank contains 100 lb of salt. Hence y(0) ⫽ 100 is the initial condition that will give the unique
solution. Substituting y ⫽ 100 and t ⫽ 0 in the last equation gives 100 ⫺ 5000 ⫽ ce0 ⫽ c. Hence c ⫽ ⫺4900.
Hence the amount of salt in the tank at time t is
y(t) ⫽ 5000 ⫺ 4900eⴚ0.01t.
(5)
This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically that
y(t) should increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE?
The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35)
or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flow
䊏
rates (in and out) may be different and known only very roughly.
y
5000
4000
3000
2000
1000
100
0
100
200
300
400
Salt content y(t)
Tank
Fig. 11. Mixing problem in Example 5
500
t
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SEC. 1.3 Separable ODEs. Modeling
EXAMPLE 6
15
Heating an Office Building (Newton’s Law of Cooling3)
Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heating
is shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M.
was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. What
was the temperature inside the building when the heat was turned on at 6 A.M.?
Physical information. Experiments show that the time rate of change of the temperature T of a body B (which
conducts heat well, for example, as a copper ball does) is proportional to the difference between T and the
temperature of the surrounding medium (Newton’s law of cooling).
Solution. Step 1. Setting up a model. Let T(t) be the temperature inside the building and TA the outside
temperature (assumed to be constant in Newton’s law). Then by Newton’s law,
dT
⫽ k(T ⫺ TA).
dt
(6)
Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if a
model seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information.
To see how good a model is, the engineer will collect experimental data and compare them with calculations
from the model.
Step 2. General solution. We cannot solve (6) because we do not know TA, just that it varied between 50°F
and 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. We
solve (6) with the unknown function TA replaced with the average of the two known values, or 45°F. For physical
reasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M.
For constant TA ⫽ 45 (or any other constant value) the ODE (6) is separable. Separation, integration, and
taking exponents gives the general solution
dT
⫽ k dt,
T ⫺ 45
ln ƒ T ⫺ 45 ƒ ⫽ kt ⫹ c*,
*
T(t) ⫽ 45 ⫹ cekt
(c ⫽ ec ).
Step 3. Particular solution. We choose 10 P.M. to be t ⫽ 0. Then the given initial condition is T(0) ⫽ 70 and
yields a particular solution, call it Tp. By substitution,
T(0) ⫽ 45 ⫹ ce0 ⫽ 70,
c ⫽ 70 ⫺ 45 ⫽ 25,
Tp(t) ⫽ 45 ⫹ 25ekt.
Step 4. Determination of k. We use T(4) ⫽ 65, where t ⫽ 4 is 2 A.M. Solving algebraically for k and inserting
k into Tp(t) gives (Fig. 12)
Tp(4) ⫽ 45 ⫹ 25e4k ⫽ 65,
e4k ⫽ 0.8,
k ⫽ 14 ln 0.8 ⫽ ⫺0.056,
Tp(t) ⫽ 45 ⫹ 25eⴚ0.056t.
y
70
68
66
65
64
62
61
60
0
2
4
6
8
t
Fig. 12. Particular solution (temperature) in Example 6
3
Sir ISAAC NEWTON (1642–1727), great English physicist and mathematician, became a professor at
Cambridge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopher
GOTTFRIED WILHELM LEIBNIZ (1646–1716) invented (independently) the differential and integral calculus.
Newton discovered many basic physical laws and created the method of investigating physical problems by
means of calculus. His Philosophiae naturalis principia mathematica (Mathematical Principles of Natural
Philosophy, 1687) contains the development of classical mechanics. His work is of greatest importance to both
mathematics and physics.
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CHAP. 1 First-Order ODEs
Step 5. Answer and interpretation. 6 A.M. is t ⫽ 8 (namely, 8 hours after 10 P.M.), and
Tp(8) ⫽ 45 ⫹ 25eⴚ0.056 ⴢ 8 ⫽ 613°F4.
䊏
Hence the temperature in the building dropped 9°F, a result that looks reasonable.
EXAMPLE 7
Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law)
This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from a
cylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank at
any time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when the
hole is opened is 2.25 m. When will the tank be empty?
Physical information. Under the influence of gravity the outflowing water has velocity
v(t) ⫽ 0.600 22gh(t)
(7)
(Torricelli’s law4),
where h(t) is the height of the water above the hole at time t, and g ⫽ 980 cm>sec2 ⫽ 32.17 ft>sec2 is the
acceleration of gravity at the surface of the earth.
Solution.
Step 1. Setting up the model. To get an equation, we relate the decrease in water level h(t) to the
outflow. The volume ¢V of the outflow during a short time ¢t is
¢V ⫽ Av ¢t
(A ⫽ Area of hole).
¢V must equal the change ¢V* of the volume of the water in the tank. Now
¢V* ⫽ ⫺B ¢h
(B ⫽ Cross-sectional area of tank)
where ¢h (⬎ 0) is the decrease of the height h(t) of the water. The minus sign appears because the volume of
the water in the tank decreases. Equating ¢V and ¢V* gives
⫺B ¢h ⫽ Av ¢t.
We now express v according to Torricelli’s law and then let ¢t (the length of the time interval considered)
approach 0—this is a standard way of obtaining an ODE as a model. That is, we have
¢h
A
A
⫽ ⫺ v ⫽ ⫺ 0.600 12gh(t)
¢t
B
B
and by letting ¢t : 0 we obtain the ODE
dh
A
⫽ ⫺26.56 1h,
dt
B
where 26.56 ⫽ 0.60022 ⴢ 980. This is our model, a first-order ODE.
Step 2. General solution. Our ODE is separable. A>B is constant. Separation and integration gives
dh
A
⫽ ⫺26.56 dt
B
1h
and
2 1h ⫽ c* ⫺ 26.56
A
t.
B
Dividing by 2 and squaring gives h ⫽ (c ⫺ 13.28At>B)2. Inserting 13.28A>B ⫽ 13.28 ⴢ 0.52p>1002p ⫽ 0.000 332
yields the general solution
h(t) ⫽ (c ⫺ 0.000 332t)2.
4
EVANGELISTA TORRICELLI (1608–1647), Italian physicist, pupil and successor of GALILEO GALILEI
(1564–1642) at Florence. The “contraction factor” 0.600 was introduced by J. C. BORDA in 1766 because the
stream has a smaller cross section than the area of the hole.
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SEC. 1.3 Separable ODEs. Modeling
17
Step 3. Particular solution. The initial height (the initial condition) is h(0) ⫽ 225 cm. Substitution of t ⫽ 0
and h ⫽ 225 gives from the general solution c2 ⫽ 225, c ⫽ 15.00 and thus the particular solution (Fig. 13)
h p(t) ⫽ (15.00 ⫺ 0.000 332t)2.
Step 4. Tank empty. h p(t) ⫽ 0 if t ⫽ 15.00>0.000 332 ⫽ 45,181 c sec d ⫽ 12.6 [hours].
Here you see distinctly the importance of the choice of units—we have been working with the cgs system,
in which time is measured in seconds! We used g ⫽ 980 cm>sec2.
䊏
Step 5. Checking. Check the result.
h
250
2.00 m
Water level
at time t
200
150
2.25 m
h(t)
100
50
Outflowing
water
0
0
10000
30000
50000
t
Water level h(t) in tank
Tank
Fig. 13. Example 7. Outflow from a cylindrical tank (“leaking tank”).
Torricelli’s law
Extended Method: Reduction to Separable Form
Certain nonseparable ODEs can be made separable by transformations that introduce for
y a new unknown function. We discuss this technique for a class of ODEs of practical
importance, namely, for equations
y
yr ⫽ f a b .
x
(8)
Here, f is any (differentiable) function of y>x, such as sin(y>x), (y>x)4, and so on. (Such
an ODE is sometimes called a homogeneous ODE, a term we shall not use but reserve
for a more important purpose in Sec. 1.5.)
The form of such an ODE suggests that we set y>x ⫽ u; thus,
(9)
y ⫽ ux
and by product differentiation
y r ⫽ u r x ⫹ u.
Substitution into y r ⫽ f (y>x) then gives u r x ⫹ u ⫽ f (u) or u r x ⫽ f (u) ⫺ u. We see that
if f (u) ⫺ u ⫽ 0, this can be separated:
(10)
dx
du
⫽ .
x
f (u) ⫺ u
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CHAP. 1 First-Order ODEs
EXAMPLE 8
Reduction to Separable Form
Solve
2xyy r ⫽ y 2 ⫺ x 2.
Solution.
To get the usual explicit form, divide the given equation by 2xy,
yr ⫽
y2 ⫺ x 2
2xy
⫽
y
2x
x
.
2y
⫺
Now substitute y and y r from (9) and then simplify by subtracting u on both sides,
urx ⫹ u ⫽
u
1
⫺ ,
2
2u
urx ⫽ ⫺
u
1
⫺u 2 ⫺ 1
⫺
⫽
.
2
2u
2u
You see that in the last equation you can now separate the variables,
2u du
1⫹u
2
⫽⫺
dx
.
x
By integration,
1
ln (1 ⫹ u 2) ⫽ ⫺ln ƒ x ƒ ⫹ c* ⫽ ln ` ` ⫹ c*.
x
Take exponents on both sides to get 1 ⫹ u 2 ⫽ c>x or 1 ⫹ (y>x)2 ⫽ c>x. Multiply the last equation by x 2 to
obtain (Fig. 14)
c 2
c2
Thus
x 2 ⫹ y 2 ⫽ cx.
ax ⫺ b ⫹ y 2 ⫽ .
2
4
This general solution represents a family of circles passing through the origin with centers on the x-axis.
䊏
y
4
2
–8
–4
4
8
x
–2
–4
Fig. 14. General solution (family of circles) in Example 8
PROBLEM SET 1.3
1. CAUTION! Constant of integration. Why is it
important to introduce the constant of integration
immediately when you integrate?
2–10
GENERAL SOLUTION
Find a general solution. Show the steps of derivation. Check
your answer by substitution.
2. y 3y r ⫹ x 3 ⫽ 0
3. y r ⫽ sec 2 y
4. y r sin 2 px ⫽ py cos 2 px
5. yy r ⫹ 36x ⫽ 0
6. y r ⫽ e2x⫺1y 2
y
7. xy r ⫽ y ⫹ 2x 3 sin 2
(Set y>x ⫽ u)
x
8. y r ⫽ (y ⫹ 4x)2 (Set y ⫹ 4x ⫽ v)
9. xy r ⫽ y 2 ⫹ y (Set y>x ⫽ u)
10. xy r ⫽ x ⫹ y (Set y>x ⫽ u)
11–17
INITIAL VALUE PROBLEMS (IVPS)
Solve the IVP. Show the steps of derivation, beginning with
the general solution.
11. xy r ⫹ y ⫽ 0,
y(4) ⫽ 6
12. y r ⫽ 1 ⫹ 4y ,
y(1) ⫽ 0
2
13. y r cosh x ⫽ sin y,
2
2
y(0) ⫽ 12 p
14. dr>dt ⫽ ⫺2tr,
r(0) ⫽ r0
15. y r ⫽ ⫺4x>y,
y(2) ⫽ 3
16. y r ⫽ (x ⫹ y ⫺ 2)2, y(0) ⫽ 2
(Set v ⫽ x ⫹ y ⫺ 2)
17. xy r ⫽ y ⫹ 3x 4 cos 2 (y>x),
(Set y>x ⫽ u)
y(1) ⫽ 0
18. Particular solution. Introduce limits of integration in
(3) such that y obtained from (3) satisfies the initial
condition y(x 0) ⫽ y0.
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SEC. 1.3 Separable ODEs. Modeling
19–36
MODELING, APPLICATIONS
19. Exponential growth. If the growth rate of the number
of bacteria at any time t is proportional to the number
present at t and doubles in 1 week, how many bacteria
can be expected after 2 weeks? After 4 weeks?
20. Another population model.
(a) If the birth rate and death rate of the number of
bacteria are proportional to the number of bacteria
present, what is the population as a function of time.
(b) What is the limiting situation for increasing time?
Interpret it.
21. Radiocarbon dating. What should be the 146 C content
(in percent of y0) of a fossilized tree that is claimed to
be 3000 years old? (See Example 4.)
22. Linear accelerators are used in physics for
accelerating charged particles. Suppose that an alpha
particle enters an accelerator and undergoes a constant
acceleration that increases the speed of the particle
from 10 3 m>sec to 10 4 m>sec in 10 ⴚ3 sec. Find the
acceleration a and the distance traveled during that
period of 10 ⴚ3 sec.
23. Boyle–Mariotte’s law for ideal gases.5 Experiments
show for a gas at low pressure p (and constant
temperature) the rate of change of the volume V(p)
equals ⫺V>p. Solve the model.
24. Mixing problem. A tank contains 400 gal of brine
in which 100 lb of salt are dissolved. Fresh water runs
into the tank at a rate of 2 gal>min.The mixture, kept
practically uniform by stirring, runs out at the same
rate. How much salt will there be in the tank at the
end of 1 hour?
25. Newton’s law of cooling. A thermometer, reading
5°C, is brought into a room whose temperature is 22°C.
One minute later the thermometer reading is 12°C.
How long does it take until the reading is practically
22°C, say, 21.9°C?
26. Gompertz growth in tumors. The Gompertz model
is y r ⫽ ⫺Ay ln y (A ⬎ 0), where y(t) is the mass of
tumor cells at time t. The model agrees well with
clinical observations. The declining growth rate with
increasing y ⬎ 1 corresponds to the fact that cells in
the interior of a tumor may die because of insufficient
oxygen and nutrients. Use the ODE to discuss the
growth and decline of solutions (tumors) and to find
constant solutions. Then solve the ODE.
27. Dryer. If a wet sheet in a dryer loses its moisture at
a rate proportional to its moisture content, and if it
loses half of its moisture during the first 10 min of
19
drying, when will it be practically dry, say, when will
it have lost 99% of its moisture? First guess, then
calculate.
28. Estimation. Could you see, practically without calculation, that the answer in Prob. 27 must lie between
60 and 70 min? Explain.
29. Alibi? Jack, arrested when leaving a bar, claims that
he has been inside for at least half an hour (which
would provide him with an alibi). The police check
the water temperature of his car (parked near the
entrance of the bar) at the instant of arrest and again
30 min later, obtaining the values 190°F and 110°F,
respectively. Do these results give Jack an alibi?
(Solve by inspection.)
30. Rocket. A rocket is shot straight up from the earth,
with a net acceleration (⫽ acceleration by the rocket
engine minus gravitational pullback) of 7t m>sec2
during the initial stage of flight until the engine cut out
at t ⫽ 10 sec. How high will it go, air resistance
neglected?
31. Solution curves of y r ⴝ g1y>x2. Show that any
(nonvertical) straight line through the origin of the
xy-plane intersects all these curves of a given ODE at
the same angle.
32. Friction. If a body slides on a surface, it experiences
friction F (a force against the direction of motion).
Experiments show that ƒ F ƒ ⫽ ␮ ƒ N ƒ (Coulomb’s6 law of
kinetic friction without lubrication), where N is the
normal force (force that holds the two surfaces together;
see Fig. 15) and the constant of proportionality ␮ is
called the coefficient of kinetic friction. In Fig. 15
assume that the body weighs 45 nt (about 10 lb; see
front cover for conversion). ␮ ⫽ 0.20 (corresponding
to steel on steel), a ⫽ 30°, the slide is 10 m long, the
initial velocity is zero, and air resistance is
negligible. Find the velocity of the body at the end
of the slide.
s(t)
Body
v(t)
N
α
W
Fig. 15. Problem 32
5
ROBERT BOYLE (1627–1691), English physicist and chemist, one of the founders of the Royal Society. EDME MARIOTTE (about
1620–1684), French physicist and prior of a monastry near Dijon. They found the law experimentally in 1662 and 1676, respectively.
6
CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer.
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CHAP. 1 First-Order ODEs
33. Rope. To tie a boat in a harbor, how many times
must a rope be wound around a bollard (a vertical
rough cylindrical post fixed on the ground) so that a
man holding one end of the rope can resist a force
exerted by the boat 1000 times greater than the man
can exert? First guess. Experiments show that the
change ¢S of the force S in a small portion of the
rope is proportional to S and to the small angle ¢␾
in Fig. 16. Take the proportionality constant 0.15.
The result should surprise you!
S
Small
portion
of rope
Δ␾
S + ΔS
Fig. 16. Problem 33
34. TEAM PROJECT. Family of Curves. A family of
curves can often be characterized as the general
solution of y r ⫽ f (x, y).
(a) Show that for the circles with center at the origin
we get y r ⫽ ⫺x>y.
(b) Graph some of the hyperbolas xy ⫽ c. Find an
ODE for them.
(c) Find an ODE for the straight lines through the
origin.
(d) You will see that the product of the right sides of
the ODEs in (a) and (c) equals ⫺1. Do you recognize
this as the condition for the two families to be
orthogonal (i.e., to intersect at right angles)? Do your
graphs confirm this?
(e) Sketch families of curves of your own choice and
find their ODEs. Can every family of curves be given
by an ODE?
35. CAS PROJECT. Graphing Solutions. A CAS can
usually graph solutions, even if they are integrals that
cannot be evaluated by the usual analytical methods of
calculus.
(a) Show2 this for the five initial value problems
y r ⫽ eⴚx , y(0) ⫽ 0, ⫾1, ⫾2, graphing all five curves
on the same axes.
(b) Graph approximate solution curves, using the first
few terms of the Maclaurin series (obtained by termwise integration of that of y r ) and compare with the
exact curves.
(c) Repeat the work in (a) for another ODE and initial
conditions of your own choice, leading to an integral
that cannot be evaluated as indicated.
36. TEAM PROJECT. Torricelli’s Law. Suppose that
the tank in Example 7 is hemispherical, of radius R,
initially full of water, and has an outlet of 5 cm2 crosssectional area at the bottom. (Make a sketch.) Set
up the model for outflow. Indicate what portion of
your work in Example 7 you can use (so that it can
become part of the general method independent of the
shape of the tank). Find the time t to empty the tank
(a) for any R, (b) for R ⫽ 1 m. Plot t as function of
R. Find the time when h ⫽ R>2 (a) for any R, (b) for
R ⫽ 1 m.
1.4 Exact ODEs. Integrating Factors
We recall from calculus that if a function u(x, y) has continuous partial derivatives, its
differential (also called its total differential) is
du ⫽
0u
0u
dx ⫹
dy.
0x
0y
From this it follows that if u(x, y) ⫽ c ⫽ const, then du ⫽ 0.
For example, if u ⫽ x ⫹ x 2y 3 ⫽ c, then
du ⫽ (1 ⫹ 2xy 3) dx ⫹ 3x 2y 2 dy ⫽ 0
or
yr ⫽
dy
1 ⫹ 2xy 3
⫽⫺
,
dx
3x 2y 2
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SEC. 1.4 Exact ODEs. Integrating Factors
21
an ODE that we can solve by going backward. This idea leads to a powerful solution
method as follows.
A first-order ODE M(x, y) ⫹ N(x, y)y r ⫽ 0, written as (use dy ⫽ y r dx as in Sec. 1.3)
M(x, y) dx ⫹ N(x, y) dy ⫽ 0
(1)
is called an exact differential equation if the differential form M(x, y) dx ⫹ N(x, y) dy
is exact, that is, this form is the differential
du ⫽
(2)
0u
0u
dx ⫹
dy
0x
0y
of some function u(x, y). Then (1) can be written
du ⫽ 0.
By integration we immediately obtain the general solution of (1) in the form
u(x, y) ⫽ c.
(3)
This is called an implicit solution, in contrast to a solution y ⫽ h(x) as defined in Sec.
1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solution
can be converted to explicit form. (Do this for x 2 ⫹ y 2 ⫽ 1.) If this is not possible, your
CAS may graph a figure of the contour lines (3) of the function u(x, y) and help you in
understanding the solution.
Comparing (1) and (2), we see that (1) is an exact differential equation if there is some
function u(x, y) such that
(4)
(a)
0u
⫽ M,
0x
(b)
0u
⫽ N.
0y
From this we can derive a formula for checking whether (1) is exact or not, as follows.
Let M and N be continuous and have continuous first partial derivatives in a region in
the xy-plane whose boundary is a closed curve without self-intersections. Then by partial
differentiation of (4) (see App. 3.2 for notation),
0M
0 2u
⫽
,
0y
0y 0x
0 2u
0N
⫽
.
0x
0x 0y
By the assumption of continuity the two second partial derivaties are equal. Thus
(5)
0N
0M
⫽
.
0y
0x
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22
Page 22
CHAP. 1 First-Order ODEs
This condition is not only necessary but also sufficient for (1) to be an exact differential
equation. (We shall prove this in Sec. 10.2 in another context. Some calculus books, for
instance, [GenRef 12], also contain a proof.)
If (1) is exact, the function u(x, y) can be found by inspection or in the following
systematic way. From (4a) we have by integration with respect to x
u⫽
(6)
冮 M dx ⫹ k(y);
in this integration, y is to be regarded as a constant, and k(y) plays the role of a “constant”
of integration. To determine k(y), we derive 0u>0y from (6), use (4b) to get dk>dy, and
integrate dk>dy to get k. (See Example 1, below.)
Formula (6) was obtained from (4a). Instead of (4a) we may equally well use (4b).
Then, instead of (6), we first have by integration with respect to y
u⫽
(6*)
冮 N dy ⫹ l(x).
To determine l(x), we derive 0u>0x from (6*), use (4a) to get dl>dx, and integrate. We
illustrate all this by the following typical examples.
EXAMPLE 1
An Exact ODE
Solve
cos (x ⫹ y) dx ⫹ (3y 2 ⫹ 2y ⫹ cos (x ⫹ y)) dy ⫽ 0.
(7)
Solution.
Step 1. Test for exactness. Our equation is of the form (1) with
M ⫽ cos (x ⫹ y),
N ⫽ 3y 2 ⫹ 2y ⫹ cos (x ⫹ y).
Thus
0M
⫽ ⫺sin (x ⫹ y),
0y
0N
⫽ ⫺sin (x ⫹ y).
0x
From this and (5) we see that (7) is exact.
Step 2. Implicit general solution. From (6) we obtain by integration
(8)
u⫽
冮 M dx ⫹ k(y) ⫽ 冮 cos (x ⫹ y) dx ⫹ k(y) ⫽ sin (x ⫹ y) ⫹ k(y).
To find k(y), we differentiate this formula with respect to y and use formula (4b), obtaining
0u
dk
⫽ cos (x ⫹ y) ⫹
⫽ N ⫽ 3y 2 ⫹ 2y ⫹ cos (x ⫹ y).
0y
dy
Hence dk>dy ⫽ 3y 2 ⫹ 2y. By integration, k ⫽ y 3 ⫹ y 2 ⫹ c*. Inserting this result into (8) and observing (3),
we obtain the answer
u(x, y) ⫽ sin (x ⫹ y) ⫹ y 3 ⫹ y 2 ⫽ c.
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SEC. 1.4 Exact ODEs. Integrating Factors
23
Step 3. Checking an implicit solution. We can check by differentiating the implicit solution u(x, y) ⫽ c
implicitly and see whether this leads to the given ODE (7):
(9)
du ⫽
0u
0u
dx ⫹
dy ⫽ cos (x ⫹ y) dx ⫹ (cos (x ⫹ y) ⫹ 3y 2 ⫹ 2y) dy ⫽ 0.
0x
0y
䊏
This completes the check.
EXAMPLE 2
An Initial Value Problem
Solve the initial value problem
(cos y sinh x ⫹ 1) dx ⫺ sin y cosh x dy ⫽ 0,
(10)
Solution.
y(1) ⫽ 2.
You may verify that the given ODE is exact. We find u. For a change, let us use (6*),
冮
u ⫽ ⫺ sin y cosh x dy ⫹ l(x) ⫽ cos y cosh x ⫹ l(x).
From this, 0u> 0x ⫽ cos y sinh x ⫹ dl>dx ⫽ M ⫽ cos y sinh x ⫹ 1. Hence dl>dx ⫽ 1. By integration, l(x) ⫽ x ⫹ c*.
This gives the general solution u(x, y) ⫽ cos y cosh x ⫹ x ⫽ c. From the initial condition, cos 2 cosh 1 ⫹ 1 ⫽
0.358 ⫽ c. Hence the answer is cos y cosh x ⫹ x ⫽ 0.358. Figure 17 shows the particular solutions for c ⫽ 0, 0.358
(thicker curve), 1, 2, 3. Check that the answer satisfies the ODE. (Proceed as in Example 1.) Also check that the
initial condition is satisfied.
䊏
y
2.5
2.0
1.5
1.0
0.5
0
0.5
1.0
1.5
2.0
2.5
3.0
x
Fig. 17. Particular solutions in Example 2
EXAMPLE 3
WARNING! Breakdown in the Case of Nonexactness
The equation ⫺y dx ⫹ x dy ⫽ 0 is not exact because M ⫽ ⫺y and N ⫽ x, so that in (5), 0M> 0y ⫽ ⫺1 but
0N> 0x ⫽ 1. Let us show that in such a case the present method does not work. From (6),
u⫽
冮 M dx ⫹ k(y) ⫽ ⫺xy ⫹ k(y),
hence
dk
0u
⫽ ⫺x ⫹ .
0y
dy
Now, 0u> 0y should equal N ⫽ x, by (4b). However, this is impossible because k(y) can depend only on y. Try
䊏
(6*); it will also fail. Solve the equation by another method that we have discussed.
Reduction to Exact Form. Integrating Factors
The ODE in Example 3 is ⫺y dx ⫹ x dy ⫽ 0. It is not exact. However, if we multiply it
by 1>x 2, we get an exact equation [check exactness by (5)!],
(11)
⫺y dx ⫹ x dy
x
2
y
y
1
⫽ ⫺ 2 dx ⫹ dy ⫽ d a b ⫽ 0.
x
x
x
Integration of (11) then gives the general solution y>x ⫽ c ⫽ const.
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24
Page 24
CHAP. 1 First-Order ODEs
This example gives the idea. All we did was to multiply a given nonexact equation, say,
P(x, y) dx ⫹ Q(x, y) dy ⫽ 0,
(12)
by a function F that, in general, will be a function of both x and y. The result was an equation
FP dx ⫹ FQ dy ⫽ 0
(13)
that is exact, so we can solve it as just discussed. Such a function F(x, y) is then called
an integrating factor of (12).
EXAMPLE 4
Integrating Factor
The integrating factor in (11) is F ⫽ 1>x 2. Hence in this case the exact equation (13) is
FP dx ⫹ FQ dy ⫽
⫺y dx ⫹ x dy
x
2
y
⫽ d a b ⫽ 0.
x
Solution
y
x
⫽ c.
These are straight lines y ⫽ cx through the origin. (Note that x ⫽ 0 is also a solution of ⫺y dx ⫹ x dy ⫽ 0.)
It is remarkable that we can readily find other integrating factors for the equation ⫺y dx ⫹ x dy ⫽ 0, namely,
1>y 2, 1>(xy), and 1>(x 2 ⫹ y 2), because
(14)
⫺y dx ⫹ x dy
y
2
x
⫽ d a b,
y
⫺y dx ⫹ x dy
xy
x
⫽ ⫺d aln b ,
y
⫺y dx ⫹ x dy
x ⫹y
2
2
y
⫽ d aarctan b .
x
䊏
How to Find Integrating Factors
In simpler cases we may find integrating factors by inspection or perhaps after some trials,
keeping (14) in mind. In the general case, the idea is the following.
For M dx ⫹ N dy ⫽ 0 the exactness condition (5) is 0M>0y ⫽ 0N>0x. Hence for (13),
FP dx ⫹ FQ dy ⫽ 0, the exactness condition is
0
0
(FP) ⫽
(FQ).
0y
0x
(15)
By the product rule, with subscripts denoting partial derivatives, this gives
FyP ⫹ FPy ⫽ FxQ ⫹ FQ x.
In the general case, this would be complicated and useless. So we follow the Golden Rule:
If you cannot solve your problem, try to solve a simpler one—the result may be useful
(and may also help you later on). Hence we look for an integrating factor depending only
on one variable: fortunately, in many practical cases, there are such factors, as we shall
see. Thus, let F ⫽ F(x). Then Fy ⫽ 0, and Fx ⫽ F r ⫽ dF>dx, so that (15) becomes
FPy ⫽ F r Q ⫹ FQ x.
Dividing by FQ and reshuffling terms, we have
(16)
1 dF
⫽ R,
F dx
where
R⫽
0Q
1 0P
a
⫺
b.
Q 0y
0x
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SEC. 1.4 Exact ODEs. Integrating Factors
25
This proves the following theorem.
THEOREM 1
Integrating Factor F (x)
If (12) is such that the right side R of (16) depends only on x, then (12) has an
integrating factor F ⫽ F(x), which is obtained by integrating (16) and taking
exponents on both sides.
冮
F(x) ⫽ exp R(x) dx.
(17)
Similarly, if F* ⫽ F*(y), then instead of (16) we get
(18)
1 dF*
⫽ R*,
F* dy
1 0Q
0P
a
⫺
b
P 0x
0y
R* ⫽
where
and we have the companion
THEOREM 2
Integrating Factor F* (y)
If (12) is such that the right side R* of (18) depends only on y, then (12) has an
integrating factor F* ⫽ F*(y), which is obtained from (18) in the form
冮
F*(y) ⫽ exp R*(y) dy.
(19)
EXAMPLE 5
Application of Theorems 1 and 2. Initial Value Problem
Using Theorem 1 or 2, find an integrating factor and solve the initial value problem
(ex⫹y ⫹ yey) dx ⫹ (xey ⫺ 1) dy ⫽ 0, y(0) ⫽ ⫺1
(20)
Solution.
Step 1. Nonexactness. The exactness check fails:
0P
0 x⫹y
⫽
(e
⫹ yey) ⫽ ex⫹y ⫹ ey ⫹ yey
0y
0y
0Q
but
0x
⫽
0
(xey ⫺ 1) ⫽ ey.
0x
Step 2. Integrating factor. General solution. Theorem 1 fails because R [the right side of (16)] depends on
both x and y.
R⫽
0Q
1 0P
1
a
⫺
b⫽ y
(ex⫹y ⫹ ey ⫹ yey ⫺ ey).
Q 0y
0x
xe ⫺ 1
Try Theorem 2. The right side of (18) is
R* ⫽
1 0Q
0P
1
(ey ⫺ ex⫹y ⫺ ey ⫺ yey) ⫽ ⫺1.
a
⫺
b ⫽ x⫹y
P 0x
0y
e
⫹ yey
Hence (19) gives the integrating factor F*(y) ⫽ eⴚy. From this result and (20) you get the exact equation
(ex ⫹ y) dx ⫹ (x ⫺ eⴚy) dy ⫽ 0.
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CHAP. 1 First-Order ODEs
Test for exactness; you will get 1 on both sides of the exactness condition. By integration, using (4a),
u⫽
冮 (e ⫹ y) dx ⫽ e ⫹ xy ⫹ k(y).
x
x
Differentiate this with respect to y and use (4b) to get
dk
0u
⫽x⫹
⫽ N ⫽ x ⫺ eⴚy,
0y
dy
dk
⫽ ⫺eⴚy,
dy
k ⫽ eⴚy ⫹ c*.
Hence the general solution is
u(x, y) ⫽ ex ⫹ xy ⫹ eⴚy ⫽ c.
Setp 3. Particular solution. The initial condition y(0) ⫽ ⫺1 gives u(0, ⫺1) ⫽ 1 ⫹ 0 ⫹ e ⫽ 3.72. Hence the
answer is ex ⫹ xy ⫹ eⴚy ⫽ 1 ⫹ e ⫽ 3.72. Figure 18 shows several particular solutions obtained as level curves
of u(x, y) ⫽ c, obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast a
solution into explicit form. Note the curve that (nearly) satisfies the initial condition.
Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initial
condition.
䊏
y
3
2
1
–3
–2
–1
0
1
2
3
x
–1
–2
–3
Fig. 18.
Particular solutions in Example 5
PROBLEM SET 1.4
1–14
ODEs. INTEGRATING FACTORS
Test for exactness. If exact, solve. If not, use an integrating
factor as given or obtained by inspection or by the theorems
in the text. Also, if an initial condition is given, find the
corresponding particular solution.
1. 2xy dx ⫹ x 2 dy ⫽ 0
2. x 3dx ⫹ y 3dy ⫽ 0
3. sin x cos y dx ⫹ cos x sin y dy ⫽ 0
4. e3u(dr ⫹ 3r du) ⫽ 0
5. (x 2 ⫹ y 2) dx ⫺ 2xy dy ⫽ 0
6. 3(y ⫹ 1) dx ⫽ 2x dy, (y ⫹ 1)x ⴚ4
7. 2x tan y dx ⫹ sec 2 y dy ⫽ 0
8. ex(cos y dx ⫺ sin y dy) ⫽ 0
9. e2x(2 cos y dx ⫺ sin y dy) ⫽ 0,
10. y dx ⫹ 3y ⫹ tan (x ⫹ y)4 dy ⫽ 0,
y(0) ⫽ 0
cos (x ⫹ y)
11. 2 cosh x cos y dx ⫽ sinh x sin y dy
2
12. (2xy dx ⫹ dy)ex ⫽ 0,
y(0) ⫽ 2
13. eⴚy dx ⫹ eⴚx(⫺eⴚy ⫹ 1) dy ⫽ 0,
F ⫽ ex⫹y
14. (a ⫹ 1)y dx ⫹ (b ⫹ 1)x dy ⫽ 0,
F ⫽ x ay b
y(1) ⫽ 1,
15. Exactness. Under what conditions for the constants a,
b, k, l is (ax ⫹ by) dx ⫹ (kx ⫹ ly) dy ⫽ 0 exact? Solve
the exact ODE.
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SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
16. TEAM PROJECT. Solution by Several Methods.
Show this as indicated. Compare the amount of work.
(a) ey(sinh x dx ⫹ cosh x dy) ⫽ 0 as an exact ODE
and by separation.
(b) (1 ⫹ 2x) cos y dx ⫹ dy>cos y ⫽ 0 by Theorem 2
and by separation.
(c) (x 2 ⫹ y 2) dx ⫺ 2xy dy ⫽ 0 by Theorem 1 or 2 and
by separation with v ⫽ y>x.
(d) 3x 2 y dx ⫹ 4x 3 dy ⫽ 0 by Theorems 1 and 2 and
by separation.
(e) Search the text and the problems for further ODEs
that can be solved by more than one of the methods
discussed so far. Make a list of these ODEs. Find
further cases of your own.
17. WRITING PROJECT. Working Backward.
Working backward from the solution to the problem
is useful in many areas. Euler, Lagrange, and other
great masters did it. To get additional insight into
the idea of integrating factors, start from a u(x, y) of
your choice, find du ⫽ 0, destroy exactness by
division by some F(x, y), and see what ODE’s
solvable by integrating factors you can get. Can you
proceed systematically, beginning with the simplest
F(x, y)?
27
18. CAS PROJECT. Graphing Particular Solutions.
Graph particular solutions of the following ODE,
proceeding as explained.
(21)
dy ⫺ y 2 sin x dx ⫽ 0.
(a) Show that (21) is not exact. Find an integrating
factor using either Theorem 1 or 2. Solve (21).
(b) Solve (21) by separating variables. Is this simpler
than (a)?
(c) Graph the seven particular solutions satisfying the
following initial conditions y(0) ⫽ 1, y(p>2) ⫽ ⫾12 ,
⫾23 , ⫾1 (see figure below).
(d) Which solution of (21) do we not get in (a) or (b)?
y
3
2
1
0
π
2π
3π
4π
x
–1
–2
–3
Particular solutions in CAS Project 18
1.5 Linear ODEs.
Bernoulli Equation.
Population Dynamics
Linear ODEs or ODEs that can be transformed to linear form are models of various
phenomena, for instance, in physics, biology, population dynamics, and ecology, as we
shall see. A first-order ODE is said to be linear if it can be brought into the form
(1)
y r ⫹ p(x)y ⫽ r(x),
by algebra, and nonlinear if it cannot be brought into this form.
The defining feature of the linear ODE (1) is that it is linear in both the unknown
function y and its derivative y r ⫽ dy>dx, whereas p and r may be any given functions of
x. If in an application the independent variable is time, we write t instead of x.
If the first term is f (x)y r (instead of y r ), divide the equation by f (x) to get the standard
form (1), with y r as the first term, which is practical.
For instance, y r cos x ⫹ y sin x ⫽ x is a linear ODE, and its standard form is
y r ⫹ y tan x ⫽ x sec x.
The function r(x) on the right may be a force, and the solution y(x) a displacement in
a motion or an electrical current or some other physical quantity. In engineering, r(x) is
frequently called the input, and y(x) is called the output or the response to the input (and,
if given, to the initial condition).
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CHAP. 1 First-Order ODEs
Homogeneous Linear ODE. We want to solve (1) in some interval a ⬍ x ⬍ b, call
it J, and we begin with the simpler special case that r(x) is zero for all x in J. (This is
sometimes written r(x) ⬅ 0.) Then the ODE (1) becomes
y r ⫹ p(x)y ⫽ 0
(2)
and is called homogeneous. By separating variables and integrating we then obtain
dy
⫽ ⫺p(x) dx,
y
冮
ln ƒ y ƒ ⫽ ⫺ p(x) dx ⫹ c*.
thus
Taking exponents on both sides, we obtain the general solution of the homogeneous
ODE (2),
(3)
y(x) ⫽ ceⴚ兰p(x) dx
(c ⫽ ⫾ec*
when
y ⭵ 0);
here we may also choose c ⫽ 0 and obtain the trivial solution y(x) ⫽ 0 for all x in that
interval.
Nonhomogeneous Linear ODE. We now solve (1) in the case that r(x) in (1) is not
everywhere zero in the interval J considered. Then the ODE (1) is called nonhomogeneous.
It turns out that in this case, (1) has a pleasant property; namely, it has an integrating factor
depending only on x. We can find this factor F(x) by Theorem 1 in the previous section
or we can proceed directly, as follows. We multiply (1) by F(x), obtaining
(1*)
Fy r ⫹ pFy ⫽ rF.
The left side is the derivative (Fy) r ⫽ F r y ⫹ Fy r of the product Fy if
pFy ⫽ F r y,
pF ⫽ F r .
thus
By separating variables, dF>F ⫽ p dx. By integration, writing h ⫽ 兰 p dx,
ln ƒ F ƒ ⫽ h ⫽
冮 p dx,
F ⫽ eh.
thus
With this F and h r ⫽ p, Eq. (1*) becomes
ehy r ⫹ h r ehy ⫽ ehy r ⫹ (eh) r y ⫽ (ehy) r ⫽ reh.
By integration,
ehy ⫽
冮 e r dx ⫹ c.
h
Dividing by eh, we obtain the desired solution formula
(4)
冮
y(x) ⫽ eⴚh a ehr dx ⫹ cb,
h⫽
冮 p(x) dx.
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SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
29
This reduces solving (1) to the generally simpler task of evaluating integrals. For ODEs
for which this is still difficult, you may have to use a numeric method for integrals from
Sec. 19.5 or for the ODE itself from Sec. 21.1. We mention that h has nothing to do with
h(x) in Sec. 1.1 and that the constant of integration in h does not matter; see Prob. 2.
The structure of (4) is interesting. The only quantity depending on a given initial
condition is c. Accordingly, writing (4) as a sum of two terms,
冮
y(x) ⫽ eⴚh ehr dx ⫹ ceⴚh,
(4*)
we see the following:
(5)
EXAMPLE 1
Total Output ⫽ Response to the Input r ⫹ Response to the Initial Data.
First-Order ODE, General Solution, Initial Value Problem
Solve the initial value problem
y r ⫹ y tan x ⫽ sin 2x,
Solution.
y(0) ⫽ 1.
Here p ⫽ tan x, r ⫽ sin 2x ⫽ 2 sin x cos x, and
h⫽
冮 p dx ⫽ 冮 tan x dx ⫽ ln ƒ sec x ƒ .
From this we see that in (4),
eh ⫽ sec x,
eⴚh ⫽ cos x,
ehr ⫽ (sec x)(2 sin x cos x) ⫽ 2 sin x,
and the general solution of our equation is
冮
y(x) ⫽ cos x a2 sin x dx ⫹ cb ⫽ c cos x ⫺ 2 cos2x.
From this and the initial condition, 1 ⫽ c # 1 ⫺ 2 # 12; thus c ⫽ 3 and the solution of our initial value problem
is y ⫽ 3 cos x ⫺ 2 cos2 x. Here 3 cos x is the response to the initial data, and ⫺2 cos2 x is the response to the
䊏
input sin 2x.
EXAMPLE 2
Electric Circuit
Model the RL-circuit in Fig. 19 and solve the resulting ODE for the current I(t) A (amperes), where t is
time. Assume that the circuit contains as an EMF E(t) (electromotive force) a battery of E ⫽ 48 V (volts), which
is constant, a resistor of R ⫽ 11 ⍀ (ohms), and an inductor of L ⫽ 0.1 H (henrys), and that the current is initially
zero.
Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm’s law) and
a voltage drop LI r ⫽ L dI>dt across the conductor, and the sum of these two voltage drops equals the EMF
(Kirchhoff’s Voltage Law, KVL).
Remark.
In general, KVL states that “The voltage (the electromotive force EMF) impressed on a closed
loop is equal to the sum of the voltage drops across all the other elements of the loop.” For Kirchoff’s Current
Law (KCL) and historical information, see footnote 7 in Sec. 2.9.
Solution.
(6)
According to these laws the model of the RL-circuit is LI r ⫹ RI ⫽ E(t), in standard form
Ir ⫹
E(t)
R
I⫽
.
L
L
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CHAP. 1 First-Order ODEs
We can solve this linear ODE by (4) with x ⫽ t, y ⫽ I, p ⫽ R>L, h ⫽ (R>L)t, obtaining the general solution
冮
I ⫽ eⴚ(R>L)t a e(
R>L)t E(t)
L
dt ⫹ c b.
By integration,
I ⫽ eⴚ(R>L)t a
(7)
E e1R>L2t
E
⫹ cb ⫽ ⫹ ceⴚ(R>L)t.
L R>L
R
In our case, R>L ⫽ 11>0.1 ⫽ 110 and E(t) ⫽ 48>0.1 ⫽ 480 ⫽ const; thus,
ⴚ110t
I ⫽ 48
.
11 ⫹ ce
In modeling, one often gets better insight into the nature of a solution (and smaller roundoff errors) by inserting
given numeric data only near the end. Here, the general solution (7) shows that the current approaches the limit
E>R ⫽ 48>11 faster the larger R>L is, in our case, R>L ⫽ 11>0.1 ⫽ 110, and the approach is very fast, from
below if I(0) ⬍ 48>11 or from above if I(0) ⬎ 48>11. If I(0) ⫽ 48>11, the solution is constant (48/11 A). See
Fig. 19.
The initial value I(0) ⫽ 0 gives I(0) ⫽ E>R ⫹ c ⫽ 0, c ⫽ ⫺E>R and the particular solution
I⫽
(8)
E
(1 ⫺ eⴚ(R>L)t),
R
thus
I⫽
48
(1 ⫺ eⴚ110t).
11
䊏
I (t)
8
R = 11 ⍀
6
4
E = 48 V
2
0
L = 0.1 H
Circuit
0.01
0.02
0.03
0.04
0.05
t
Current I(t)
Fig. 19. RL-circuit
EXAMPLE 3
Hormone Level
Assume that the level of a certain hormone in the blood of a patient varies with time. Suppose that the time rate
of change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuous
removal rate proportional to the level present. Set up a model for the hormone level in the blood and find its
general solution. Find the particular solution satisfying a suitable initial condition.
Solution.
Step 1. Setting up a model. Let y(t) be the hormone level at time t. Then the removal rate is Ky(t).
The input rate is A ⫹ B cos vt, where v ⫽ 2p>24 ⫽ p>12 and A is the average input rate; here A ⭌ B to make
the input rate nonnegative. The constants A, B, K can be determined from measurements. Hence the model is the
linear ODE
y r (t) ⫽ In ⫺ Out ⫽ A ⫹ B cos vt ⫺ Ky(t),
thus
y r ⫹ Ky ⫽ A ⫹ B cos vt.
The initial condition for a particular solution ypart is ypart(0) ⫽ y0 with t ⫽ 0 suitably chosen, for example,
6:00 A.M.
Step 2. General solution. In (4) we have p ⫽ K ⫽ const, h ⫽ Kt, and r ⫽ A ⫹ B cos vt. Hence (4) gives the
general solution (evaluate 兰 eKt cos vt dt by integration by parts)
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SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
31
冮
y(t) ⫽ eⴚKt eKt aA ⫹ B cos vtb dt ⫹ ceⴚKt
⫽ eⴚKteKt c
⫽
A
B
⫹ 2
aK cos vt ⫹ v sin vtb d ⫹ ceⴚKt
K
K ⫹ v2
B
A
pt p
pt
⫹ 2
aK cos
⫹
sin b ⫹ ceⴚKt.
K
12
12
12
K ⫹ (p>12)2
The last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initial
condition). The other part of y(t) is called the steady-state solution because it consists of constant and periodic
terms. The entire solution is called the transient-state solution because it models the transition from rest to the
steady state. These terms are used quite generally for physical and other systems whose behavior depends on time.
Step 3. Particular solution. Setting t ⫽ 0 in y(t) and choosing y0 ⫽ 0, we have
y(0) ⫽
A
B
u
⫹ 2
K ⫹ c ⫽ 0,
K
K ⫹ (p>12)2 p
thus
c⫽⫺
A
KB
⫺ 2
.
K
K ⫹ (p>12)2
Inserting this result into y(t), we obtain the particular solution
ypart(t) ⫽
A
B
pt p
pt
A
KB
⫹ 2
aK cos
⫹
sin b ⫺ a ⫹ 2
b eⴚK
K
12
12
12
K
K ⫹ (p>12)2
K ⫹ (p>12)2
with the steady-state part as before. To plot ypart we must specify values for the constants, say, A ⫽ B ⫽ 1
and K ⫽ 0.05. Figure 20 shows this solution. Notice that the transition period is relatively short (although
1
pt) ⫽
K is small), and the curve soon looks sinusoidal; this is the response to the input A ⫹ B cos (12
1
1 ⫹ cos (12
pt).
䊏
y
25
20
15
10
5
0
0
100
200
t
Fig. 20. Particular solution in Example 3
Reduction to Linear Form. Bernoulli Equation
Numerous applications can be modeled by ODEs that are nonlinear but can be transformed
to linear ODEs. One of the most useful ones of these is the Bernoulli equation7
(9)
y r ⫹ p(x)y ⫽ g(x)y a
(a any real number).
7
JAKOB BERNOULLI (1654–1705), Swiss mathematician, professor at Basel, also known for his contribution
to elasticity theory and mathematical probability. The method for solving Bernoulli’s equation was discovered by
Leibniz in 1696. Jakob Bernoulli’s students included his nephew NIKLAUS BERNOULLI (1687–1759), who
contributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI (1667–1748),
who had profound influence on the development of calculus, became Jakob’s successor at Basel, and had among
his students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIEL
BERNOULLI (1700–1782) is known for his basic work in fluid flow and the kinetic theory of gases.
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CHAP. 1 First-Order ODEs
If a ⫽ 0 or a ⫽ 1, Equation (9) is linear. Otherwise it is nonlinear. Then we set
u(x) ⫽ 3y(x)41ⴚa.
We differentiate this and substitute y r from (9), obtaining
u r ⫽ (1 ⫺ a)y ⴚay r ⫽ (1 ⫺ a)y ⴚa(gy a ⫺ py).
Simplification gives
u r ⫽ (1 ⫺ a)(g ⫺ py 1ⴚa),
where y 1ⴚa ⫽ u on the right, so that we get the linear ODE
u r ⫹ (1 ⫺ a)pu ⫽ (1 ⫺ a)g.
(10)
For further ODEs reducible to linear form, see lnce’s classic [A11] listed in App. 1. See
also Team Project 30 in Problem Set 1.5.
EXAMPLE 4
Logistic Equation
Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation8):
y r ⫽ Ay ⫺ By 2
(11)
Solution.
Write (11) in the form (9), that is,
y r ⫺ Ay ⫽ ⫺By 2
to see that a ⫽ 2, so that u ⫽ y 1ⴚa ⫽ y ⴚ1. Differentiate this u and substitute y r from (11),
u r ⫽ ⫺y ⴚ2y r ⫽ ⫺y ⴚ2(Ay ⫺ By 2) ⫽ B ⫺ Ay ⫺1.
The last term is ⫺Ay ⴚ1 ⫽ ⫺Au. Hence we have obtained the linear ODE
u r ⫹ Au ⫽ B.
The general solution is [by (4)]
u ⫽ ceⴚAt ⫹ B>A.
Since u ⫽ 1>y, this gives the general solution of (11),
(12)
y⫽
1
1
⫽ ⴚAt
u
ce
⫹ B>A
Directly from (11) we see that y ⬅ 0 (y(t) ⫽ 0 for all t) is also a solution.
8
(Fig. 21)
䊏
PIERRE-FRANÇOIS VERHULST, Belgian statistician, who introduced Eq. (8) as a model for human
population growth in 1838.
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SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
33
Population y
8
6
A
=4
B
2
0
1
2
3
4
Time t
Fig. 21. Logistic population model. Curves (9) in Example 4 with A>B ⫽ 4
Population Dynamics
The logistic equation (11) plays an important role in population dynamics, a field
that models the evolution of populations of plants, animals, or humans over time t.
If B ⫽ 0, then (11) is y r ⫽ dy>dt ⫽ Ay. In this case its solution (12) is y ⫽ (1>c)eAt
and gives exponential growth, as for a small population in a large country (the
United States in early times!). This is called Malthus’s law. (See also Example 3 in
Sec. 1.1.)
The term ⫺By 2 in (11) is a “braking term” that prevents the population from growing
without bound. Indeed, if we write y r ⫽ Ay 31 ⫺ (B>A)y4, we see that if y ⬍ A>B, then
y r ⬎ 0, so that an initially small population keeps growing as long as y ⬍ A>B. But if
y ⬎ A>B, then y r ⬍ 0 and the population is decreasing as long as y ⬎ A>B. The limit
is the same in both cases, namely, A>B. See Fig. 21.
We see that in the logistic equation (11) the independent variable t does not occur
explicitly. An ODE y r ⫽ f (t, y) in which t does not occur explicitly is of the form
(13)
y r ⫽ f (y)
and is called an autonomous ODE. Thus the logistic equation (11) is autonomous.
Equation (13) has constant solutions, called equilibrium solutions or equilibrium
points. These are determined by the zeros of f (y), because f (y) ⫽ 0 gives y r ⫽ 0 by
(13); hence y ⫽ const. These zeros are known as critical points of (13). An
equilibrium solution is called stable if solutions close to it for some t remain close
to it for all further t. It is called unstable if solutions initially close to it do not remain
close to it as t increases. For instance, y ⫽ 0 in Fig. 21 is an unstable equilibrium
solution, and y ⫽ 4 is a stable one. Note that (11) has the critical points y ⫽ 0 and
y ⫽ A>B.
EXAMPLE 5
Stable and Unstable Equilibrium Solutions. “Phase Line Plot”
The ODE y r ⫽ (y ⫺ 1)(y ⫺ 2) has the stable equilibrium solution y1 ⫽ 1 and the unstable y2 ⫽ 2, as the direction
field in Fig. 22 suggests. The values y1 and y2 are the zeros of the parabola f (y) ⫽ (y ⫺ 1)(y ⫺ 2) in the figure.
Now, since the ODE is autonomous, we can “condense” the direction field to a “phase line plot” giving y1 and
y2, and the direction (upward or downward) of the arrows in the field, and thus giving information about the
stability or instability of the equilibrium solutions.
䊏
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CHAP. 1 First-Order ODEs
y(x)
y
3.0
2.0
2.5
1.5
y2
y2
2.0
1.0
1.5
y1
y1
1.0
0.5
0.5
y2
y1
–2
–1
0
1
2
(a)
Fig. 22.
x
0
(b)
0.5
1.0
1.5
2.0
2.5
3.0
x
(c)
Example 5. (A) Direction field. (B) “Phase line”. (C) Parabola f (y)
A few further population models will be discussed in the problem set. For some more
details of population dynamics, see C. W. Clark. Mathematical Bioeconomics: The
Mathematics of Conservation 3rd ed. Hoboken, NJ, Wiley, 2010.
Further applications of linear ODEs follow in the next section.
PROBLEM SET 1.5
1. CAUTION! Show that eⴚln x ⫽ 1>x (not ⫺x) and
eⴚln(sec x) ⫽ cos x.
2. Integration constant. Give a reason why in (4) you may
choose the constant of integration in 兰 p dx to be zero.
GENERAL SOLUTION. INITIAL VALUE
3–13
PROBLEMS
Find the general solution. If an initial condition is given,
find also the corresponding particular solution and graph or
sketch it. (Show the details of your work.)
3. y r ⫺ y ⫽ 5.2
4. y r ⫽ 2y ⫺ 4x
5. y r ⫹ ky ⫽ eⴚkx
6. y r ⫹ 2y ⫽ 4 cos 2x, y(14 p) ⫽ 3
7. xy r ⫽ 2y ⫹ x 3ex
8. y r ⫹ y tan x ⫽ eⴚ0.01x cos x, y(0) ⫽ 0
9. y r ⫹ y sin x ⫽ ecos x, y(0) ⫽ ⫺2.5
10. y r cos x ⫹ (3y ⫺ 1) sec x ⫽ 0, y(14 p) ⫽ 4>3
11. y r ⫽ (y ⫺ 2) cot x
12. xy r ⫹ 4y ⫽ 8x 4, y(1) ⫽ 2
13. y r ⫽ 6(y ⫺ 2.5) tanh 1.5x
14. CAS EXPERIMENT. (a) Solve the ODE y r ⫺ y>x ⫽
⫺x ⴚ1 cos (1>x). Find an initial condition for which the
arbitrary constant becomes zero. Graph the resulting
particular solution, experimenting to obtain a good
figure near x ⫽ 0.
(b) Generalizing (a) from n ⫽ 1 to arbitrary n, solve the
ODE y r ⫺ ny>x ⫽ ⫺x nⴚ2 cos (1>x). Find an initial
condition as in (a) and experiment with the graph.
15–20
GENERAL PROPERTIES OF LINEAR ODEs
These properties are of practical and theoretical importance
because they enable us to obtain new solutions from given
ones. Thus in modeling, whenever possible, we prefer linear
ODEs over nonlinear ones, which have no similar properties.
Show that nonhomogeneous linear ODEs (1) and homogeneous linear ODEs (2) have the following properties.
Illustrate each property by a calculation for two or three
equations of your choice. Give proofs.
15. The sum y1 ⫹ y2 of two solutions y1 and y2 of the
homogeneous equation (2) is a solution of (2), and so is
a scalar multiple ay1 for any constant a. These properties
are not true for (1)!
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SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
16. y ⫽ 0 (that is, y(x) ⫽ 0 for all x, also written y(x) ⬅ 0)
is a solution of (2) [not of (1) if r(x) ⫽ 0!], called the
trivial solution.
17. The sum of a solution of (1) and a solution of (2) is a
solution of (1).
18. The difference of two solutions of (1) is a solution of (2).
19. If y1 is a solution of (1), what can you say about cy1?
20. If y1 and y2 are solutions of y1r ⫹ py1 ⫽ r1 and
y2r ⫹ py2 ⫽ r2, respectively (with the same p!), what
can you say about the sum y1 ⫹ y2?
21. Variation of parameter. Another method of obtaining
(4) results from the following idea. Write (3) as cy*,
where y* is the exponential function, which is a solution
of the homogeneous linear ODE y* r ⫹ py* ⫽ 0.
Replace the arbitrary constant c in (3) with a function
u to be determined so that the resulting function y ⫽ uy*
is a solution of the nonhomogeneous linear ODE
y r ⫹ py ⫽ r.
(b) Show that y ⫽ Y ⫽ x is a solution of the ODE
y r ⫺ (2x 3 ⫹ 1) y ⫽ ⫺x 2y 2 ⫺ x 4 ⫺ x ⫹ 1 and solve this
Riccati equation, showing the details.
(c) Solve the Clairaut equation y r 2 ⫺ xy r ⫹ y ⫽ 0 as
follows. Differentiate it with respect to x, obtaining
y s (2y r ⫺ x) ⫽ 0. Then solve (A) y s ⫽ 0 and (B)
2y r ⫺ x ⫽ 0 separately and substitute the two solutions
(a) and (b) of (A) and (B) into the given ODE. Thus
obtain (a) a general solution (straight lines) and (b) a
parabola for which those lines (a) are tangents (Fig. 6
in Prob. Set 1.1); so (b) is the envelope of (a). Such a
solution (b) that cannot be obtained from a general
solution is called a singular solution.
(d) Show that the Clairaut equation (15) has as
solutions a family of straight lines y ⫽ cx ⫹ g(c) and
a singular solution determined by g r (s) ⫽ ⫺x, where
s ⫽ y r , that forms the envelope of that family.
31–40
22–28
NONLINEAR ODEs
Using a method of this section or separating variables, find
the general solution. If an initial condition is given, find
also the particular solution and sketch or graph it.
22. y r ⫹ y ⫽ y 2, y(0) ⫽ ⫺13
23. y r ⫹ xy ⫽ xy ⴚ1, y(0) ⫽ 3
24. y r ⫹ y ⫽ ⫺x>y
25. y r ⫽ 3.2y ⫺ 10y 2
26. y r ⫽ (tan y)>(x ⫺ 1), y(0) ⫽ 12 p
27. y r ⫽ 1>(6ey ⫺ 2x)
28. 2xyy r ⫹ (x ⫺ 1)y 2 ⫽ x 2ex (Set y 2 ⫽ z)
29. REPORT PROJECT. Transformation of ODEs.
We have transformed ODEs to separable form, to exact
form, and to linear form. The purpose of such
transformations is an extension of solution methods to
larger classes of ODEs. Describe the key idea of each
of these transformations and give three typical examples of your choice for each transformation. Show each
step (not just the transformed ODE).
30. TEAM PROJECT. Riccati Equation. Clairaut
Equation. Singular Solution.
A Riccati equation is of the form
(14)
y r ⫹ p(x)y ⫽ g(x)y 2 ⫹ h(x).
A Clairaut equation is of the form
(15)
y ⫽ xy r ⫹ g(y r ).
(a) Apply the transformation y ⫽ Y ⫹ 1>u to the
Riccati equation (14), where Y is a solution of (14), and
obtain for u the linear ODE u r ⫹ (2Yg ⫺ p)u ⫽ ⫺g.
Explain the effect of the transformation by writing it
as y ⫽ Y ⫹ v, v ⫽ 1>u.
35
MODELING. FURTHER APPLICATIONS
31. Newton’s law of cooling. If the temperature of a cake
is 300°F when it leaves the oven and is 200°F ten
minutes later, when will it be practically equal to the
room temperature of 60°F, say, when will it be 61°F?
32. Heating and cooling of a building. Heating and
cooling of a building can be modeled by the ODE
T r ⫽ k 1(T ⫺ Ta) ⫹ k 2(T ⫺ Tv) ⫹ P,
where T ⫽ T(t) is the temperature in the building at
time t, Ta the outside temperature, Tw the temperature
wanted in the building, and P the rate of increase of T
due to machines and people in the building, and k 1 and
k 2 are (negative) constants. Solve this ODE, assuming
P ⫽ const, Tw ⫽ const, and Ta varying sinusoidally
over 24 hours, say, Ta ⫽ A ⫺ C cos(2 p>24)t. Discuss
the effect of each term of the equation on the solution.
33. Drug injection. Find and solve the model for drug
injection into the bloodstream if, beginning at t ⫽ 0, a
constant amount A g> min is injected and the drug is
simultaneously removed at a rate proportional to the
amount of the drug present at time t.
34. Epidemics. A model for the spread of contagious
diseases is obtained by assuming that the rate of spread
is proportional to the number of contacts between
infected and noninfected persons, who are assumed to
move freely among each other. Set up the model. Find
the equilibrium solutions and indicate their stability or
instability. Solve the ODE. Find the limit of the
proportion of infected persons as t : ⬁ and explain
what it means.
35. Lake Erie. Lake Erie has a water volume of about
450 km3 and a flow rate (in and out) of about 175 km2
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CHAP. 1 First-Order ODEs
per year. If at some instant the lake has pollution
concentration p ⫽ 0.04 %, how long, approximately,
will it take to decrease it to p> 2, assuming that the
inflow is much cleaner, say, it has pollution
concentration p> 4, and the mixture is uniform (an
assumption that is only imperfectly true)? First guess.
36. Harvesting renewable resources. Fishing. Suppose
that the population y(t) of a certain kind of fish is given
by the logistic equation (11), and fish are caught at a
rate Hy proportional to y. Solve this so-called Schaefer
model. Find the equilibrium solutions y1 and y2 (⬎ 0)
when H ⬍ A. The expression Y ⫽ Hy2 is called
the equilibrium harvest or sustainable yield corresponding to H. Why?
37. Harvesting. In Prob. 36 find and graph the solution
satisfying y(0) ⫽ 2 when (for simplicity) A ⫽ B ⫽ 1
and H ⫽ 0.2. What is the limit? What does it mean?
What if there were no fishing?
38. Intermittent harvesting. In Prob. 36 assume that you
fish for 3 years, then fishing is banned for the next
3 years. Thereafter you start again. And so on. This is
called intermittent harvesting. Describe qualitatively
how the population will develop if intermitting is
continued periodically. Find and graph the solution for
the first 9 years, assuming that A ⫽ B ⫽ 1, H ⫽ 0.2,
and y(0) ⫽ 2.
y
2
1.8
1.6
1.4
1.2
1
0.8
0
Fig. 23.
2
4
6
8
t
Fish population in Problem 38
39. Extinction vs. unlimited growth. If in a population
y(t) the death rate is proportional to the population, and
the birth rate is proportional to the chance encounters
of meeting mates for reproduction, what will the model
be? Without solving, find out what will eventually
happen to a small initial population. To a large one.
Then solve the model.
40. Air circulation. In a room containing 20,000 ft 3 of air,
600 ft 3of fresh air flows in per minute, and the mixture
(made practically uniform by circulating fans) is
exhausted at a rate of 600 cubic feet per minute (cfm).
What is the amount of fresh air y(t) at any time if
y(0) ⫽ 0? After what time will 90% of the air be fresh?
1.6 Orthogonal Trajectories.
Optional
An important type of problem in physics or geometry is to find a family of curves that
intersects a given family of curves at right angles. The new curves are called orthogonal
trajectories of the given curves (and conversely). Examples are curves of equal
temperature (isotherms) and curves of heat flow, curves of equal altitude (contour lines)
on a map and curves of steepest descent on that map, curves of equal potential
(equipotential curves, curves of equal voltage—the ellipses in Fig. 24) and curves of
electric force (the parabolas in Fig. 24).
Here the angle of intersection between two curves is defined to be the angle between
the tangents of the curves at the intersection point. Orthogonal is another word for
perpendicular.
In many cases orthogonal trajectories can be found using ODEs. In general, if we
consider G(x, y, c) ⫽ 0 to be a given family of curves in the xy-plane, then each value of
c gives a particular curve. Since c is one parameter, such a family is called a oneparameter family of curves.
In detail, let us explain this method by a family of ellipses
(1)
1 2
2
2 x ⫹ y ⫽ c
(c ⬎ 0)
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SEC. 1.6 Orthogonal Trajectories. Optional
37
and illustrated in Fig. 24. We assume that this family of ellipses represents electric
equipotential curves between the two black ellipses (equipotential surfaces between two
elliptic cylinders in space, of which Fig. 24 shows a cross-section). We seek the
orthogonal trajectories, the curves of electric force. Equation (1) is a one-parameter family
with parameter c. Each value of c (⬎ 0) corresponds to one of these ellipses.
Step 1. Find an ODE for which the given family is a general solution. Of course, this
ODE must no longer contain the parameter c. Differentiating (1), we have x ⫹ 2yy r ⫽ 0.
Hence the ODE of the given curves is
y r ⫽ f (x, y) ⫽ ⫺
(2)
x
.
2y
y
4
6 x
–6
–4
Fig. 24. Electrostatic field between two ellipses (elliptic cylinders in space):
Elliptic equipotential curves (equipotential surfaces) and orthogonal
trajectories (parabolas)
Step 2.
Find an ODE for the orthogonal trajectories y苲 ⫽ y苲(x). This ODE is
苲
yr ⫽ ⫺
(3)
苲
2y
1
⫽
⫹
x
f (x, 苲
y)
with the same f as in (2). Why? Well, a given curve passing through a point (x 0, y0) has
slope f (x 0, y0) at that point, by (2). The trajectory through (x 0, y0) has slope ⫺1>f (x 0, y0)
by (3). The product of these slopes is ⫺1, as we see. From calculus it is known that this
is the condition for orthogonality (perpendicularity) of two straight lines (the tangents at
(x 0, y0)), hence of the curve and its orthogonal trajectory at (x 0, y0).
Step 3.
Solve (3) by separating variables, integrating, and taking exponents:
d y苲
dx
⫽2 ,
x
y苲
ln ƒ y苲 ƒ ⫽ 2 ln x ⫹ c,
苲
y ⫽ c* x 2.
This is the family of orthogonal trajectories, the quadratic parabolas along which electrons
or other charged particles (of very small mass) would move in the electric field between
the black ellipses (elliptic cylinders).
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CHAP. 1 First-Order ODEs
PROBLEM SET 1.6
1–3
FAMILIES OF CURVES
Represent the given family of curves in the form
G(x, y; c) ⫽ 0 and sketch some of the curves.
1. All ellipses with foci ⫺3 and 3 on the x-axis.
2. All circles with centers on the cubic parabola y ⫽ x 3
and passing through the origin (0, 0).
3. The catenaries obtained by translating the catenary
y ⫽ cosh x in the direction of the straight line y ⫽ x.
4–10
Fig. 25.
ORTHOGONAL TRAJECTORIES (OTs)
Sketch or graph some of the given curves. Guess what their
OTs may look like. Find these OTs.
4. y ⫽ x 2 ⫹ c
5. y ⫽ cx
6. xy ⫽ c
7. y ⫽ c>x 2
8. y ⫽ 2x ⫹ c
9. y ⫽ ceⴚx
2
10. x 2 ⫹ (y ⫺ c)2 ⫽ c2
11–16
APPLICATIONS, EXTENSIONS
11. Electric field. Let the electric equipotential lines
(curves of constant potential) between two concentric
cylinders with the z-axis in space be given by
u(x, y) ⫽ x 2 ⫹ y 2 ⫽ c (these are circular cylinders in
the xyz-space). Using the method in the text, find their
orthogonal trajectories (the curves of electric force).
12. Electric field. The lines of electric force of two opposite
charges of the same strength at (⫺1, 0) and (1, 0) are
the circles through (⫺1, 0) and (1, 0) . Show that these
circles are given by x 2 ⫹ (y ⫺ c)2 ⫽ 1 ⫹ c2. Show
that the equipotential lines (which are orthogonal
trajectories of those circles) are the circles given by
(x ⫹ c*)2 ⫹ y苲 2 ⫽ c* 2 ⫺ 1 (dashed in Fig. 25).
Electric field in Problem 12
13. Temperature field. Let the isotherms (curves of
constant temperature) in a body in the upper half-plane
y ⬎ 0 be given by 4x 2 ⫹ 9y 2 ⫽ c. Find the orthogonal trajectories (the curves along which heat will
flow in regions filled with heat-conducting material and
free of heat sources or heat sinks).
14. Conic sections. Find the conditions under which
the orthogonal trajectories of families of ellipses
x 2>a 2 ⫹ y 2>b 2 ⫽ c are again conic sections. Illustrate
your result graphically by sketches or by using your
CAS. What happens if a : 0? If b : 0?
15. Cauchy–Riemann equations. Show that for a family
u(x, y) ⫽ c ⫽ const the orthogonal trajectories v(x, y) ⫽
c* ⫽ const can be obtained from the following
Cauchy–Riemann equations (which are basic in
complex analysis in Chap. 13) and use them to find the
orthogonal trajectories of ex sin y ⫽ const. (Here, subscripts denote partial derivatives.)
u x ⫽ vy,
u y ⫽ ⫺vx
16. Congruent OTs. If y r ⫽ f (x) with f independent of y,
show that the curves of the corresponding family are
congruent, and so are their OTs.
1.7 Existence and Uniqueness of Solutions
for Initial Value Problems
The initial value problem
ƒ y r ƒ ⫹ ƒ y ƒ ⫽ 0,
y(0) ⫽ 1
has no solution because y ⫽ 0 (that is, y(x) ⫽ 0 for all x) is the only solution of the ODE.
The initial value problem
y r ⫽ 2x,
y(0) ⫽ 1
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SEC. 1.7 Existence and Uniqueness of Solutions
39
has precisely one solution, namely, y ⫽ x 2 ⫹ 1. The initial value problem
xy r ⫽ y ⫺ 1,
y(0) ⫽ 1
has infinitely many solutions, namely, y ⫽ 1 ⫹ cx, where c is an arbitrary constant because
y(0) ⫽ 1 for all c.
From these examples we see that an initial value problem
y r ⫽ f (x, y),
(1)
y(x 0) ⫽ y0
may have no solution, precisely one solution, or more than one solution. This fact leads
to the following two fundamental questions.
Problem of Existence
Under what conditions does an initial value problem of the form (1) have at least
one solution (hence one or several solutions)?
Problem of Uniqueness
Under what conditions does that problem have at most one solution (hence excluding
the case that is has more than one solution)?
Theorems that state such conditions are called existence theorems and uniqueness
theorems, respectively.
Of course, for our simple examples, we need no theorems because we can solve these
examples by inspection; however, for complicated ODEs such theorems may be of
considerable practical importance. Even when you are sure that your physical or other
system behaves uniquely, occasionally your model may be oversimplified and may not
give a faithful picture of reality.
THEOREM 1
Existence Theorem
Let the right side f (x, y) of the ODE in the initial value problem
(1)
y r ⫽ f (x, y),
y(x 0) ⫽ y0
be continuous at all points (x, y) in some rectangle
R: ƒ x ⫺ x 0 ƒ ⬍ a,
ƒ y ⫺ y0 ƒ ⬍ b
(Fig. 26)
and bounded in R; that is, there is a number K such that
(2)
ƒ f (x, y) ƒ ⬉ K
for all (x, y) in R.
Then the initial value problem (1) has at least one solution y(x). This solution exists
at least for all x in the subinterval ƒ x ⫺ x 0 ƒ ⬍ a of the interval ƒ x ⫺ x 0 ƒ ⬍ a;
here, a is the smaller of the two numbers a and b> K.
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CHAP. 1 First-Order ODEs
y
y0 + b
R
y0
y0 – b
x0 – a
Fig. 26.
x0
x0 + a
x
Rectangle R in the existence and uniqueness theorems
(Example of Boundedness. The function f (x, y) ⫽ x 2 ⫹ y 2 is bounded (with K ⫽ 2) in the
square ƒ x ƒ ⬍ 1, ƒ y ƒ ⬍ 1. The function f (x, y) ⫽ tan (x ⫹ y) is not bounded for
ƒ x ⫹ y ƒ ⬍ p>2. Explain!)
THEOREM 2
Uniqueness Theorem
Let f and its partial derivative fy ⫽ 0f>0y be continuous for all (x, y) in the rectangle
R (Fig. 26) and bounded, say,
(3)
(a)
ƒ f (x, y) ƒ ⬉ K,
(b)
ƒ fy(x, y) ƒ ⬉ M
for all (x, y) in R.
Then the initial value problem (1) has at most one solution y(x). Thus, by Theorem 1,
the problem has precisely one solution. This solution exists at least for all x in that
subinterval ƒ x ⫺ x 0 ƒ ⬍ a.
Understanding These Theorems
These two theorems take care of almost all practical cases. Theorem 1 says that if f (x, y)
is continuous in some region in the xy-plane containing the point (x 0, y0), then the initial
value problem (1) has at least one solution.
Theorem 2 says that if, moreover, the partial derivative 0f>0y of f with respect to y
exists and is continuous in that region, then (1) can have at most one solution; hence, by
Theorem 1, it has precisely one solution.
Read again what you have just read—these are entirely new ideas in our discussion.
Proofs of these theorems are beyond the level of this book (see Ref. [A11] in App. 1);
however, the following remarks and examples may help you to a good understanding of
the theorems.
Since y r ⫽ f (x, y), the condition (2) implies that ƒ y r ƒ ⬉ K; that is, the slope of any
solution curve y(x) in R is at least ⫺K and at most K. Hence a solution curve that passes
through the point (x 0, y0) must lie in the colored region in Fig. 27 bounded by the lines
l 1 and l 2 whose slopes are ⫺K and K, respectively. Depending on the form of R, two
different cases may arise. In the first case, shown in Fig. 27a, we have b>K ⭌ a and
therefore a ⫽ a in the existence theorem, which then asserts that the solution exists for all
x between x 0 ⫺ a and x 0 ⫹ a. In the second case, shown in Fig. 27b, we have b>K ⬍ a.
Therefore, a ⫽ b>K ⬍ a, and all we can conclude from the theorems is that the solution
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SEC. 1.7 Existence and Uniqueness of Solutions
41
exists for all x between x 0 ⫺ b>K and x 0 ⫹ b>K. For larger or smaller x’s the solution
curve may leave the rectangle R, and since nothing is assumed about f outside R, nothing
can be concluded about the solution for those larger or amaller x’s; that is, for such x’s
the solution may or may not exist—we don’t know.
y
y
y0 + b
R
l1
l1
y0 + b
y0
R
y0
y0 – b
l2
l2
α
y0 – b
α=a
α=a
α
a
a
x
x0
x0
(a)
x
(b)
Fig. 27. The condition (2) of the existence theorem. (a) First case. (b) Second case
Let us illustrate our discussion with a simple example. We shall see that our choice of
a rectangle R with a large base (a long x-interval) will lead to the case in Fig. 27b.
EXAMPLE 1
Choice of a Rectangle
Consider the initial value problem
y r ⫽ 1 ⫹ y 2,
y(0) ⫽ 0
and take the rectangle R; ƒ x ƒ ⬍ 5, ƒ y ƒ ⬍ 3. Then a ⫽ 5, b ⫽ 3, and
ƒ f (x, y) ƒ ⫽ ƒ 1 ⫹ y 2 ƒ ⬉ K ⫽ 10,
`
0f
0y
` ⫽ 2 ƒ y ƒ ⬉ M ⫽ 6,
a⫽
b
⫽ 0.3 ⬍ a.
K
Indeed, the solution of the problem is y ⫽ tan x (see Sec. 1.3, Example 1). This solution is discontinuous at
⫾p>2, and there is no continuous solution valid in the entire interval ƒ x ƒ ⬍ 5 from which we started.
䊏
The conditions in the two theorems are sufficient conditions rather than necessary ones,
and can be lessened. In particular, by the mean value theorem of differential calculus we
have
f (x, y2) ⫺ f (x, y1) ⫽ (y2 ⫺ y1)
0f
`
0y y⫽y苲
y is a suitable value between y1
where (x, y1) and (x, y2) are assumed to be in R, and 苲
and y2. From this and (3b) it follows that
(4)
ƒ f (x, y2) ⫺ f (x, y1) ƒ ⬉ M ƒ y2 ⫺ y1 ƒ .
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CHAP. 1 First-Order ODEs
It can be shown that (3b) may be replaced by the weaker condition (4), which is known
as a Lipschitz condition.9 However, continuity of f (x, y) is not enough to guarantee the
uniqueness of the solution. This may be illustrated by the following example.
EXAMPLE 2
Nonuniqueness
The initial value problem
yr ⫽ 2 ƒ y ƒ .
y(0) ⫽ 0
has the two solutions
y⫽0
y* ⫽ e
and
x 2> 4 if
⫺x 2>4 if
x⭌0
x ⬍ 0
although f (x, y) ⫽ 2 ƒ y ƒ is continuous for all y. The Lipschitz condition (4) is violated in any region that includes
the line y ⫽ 0, because for y1 ⫽ 0 and positive y2 we have
(5)
ƒ f (x, y2) ⫺ f (x, y1) ƒ
ƒ y2 ⫺ y1 ƒ
⫽
2y2
y2
⫽
1
2y2
( 2y2 ⬎ 0)
,
and this can be made as large as we please by choosing y2 sufficiently small, whereas (4) requires that the
䊏
quotient on the left side of (5) should not exceed a fixed constant M.
PROBLEM SET 1.7
1. Linear ODE. If p and r in y r ⫹ p(x)y ⫽ r(x) are
continuous for all x in an interval ƒ x ⫺ x 0 ƒ ⱕ a, show
that f (x, y) in this ODE satisfies the conditions of our
present theorems, so that a corresponding initial value
problem has a unique solution. Do you actually need
these theorems for this ODE?
2. Existence? Does the initial value problem
(x ⫺ 2)y r ⫽ y, y(2) ⫽ 1 have a solution? Does your
result contradict our present theorems?
3. Vertical strip. If the assumptions of Theorems 1 and
2 are satisfied not merely in a rectangle but in a vertical
infinite strip ƒ x ⫺ x 0 ƒ ⬍ a, in what interval will the
solution of (1) exist?
4. Change of initial condition. What happens in Prob.
2 if you replace y(2) ⫽ 1 with y(2) ⫽ k?
5. Length of x-interval. In most cases the solution of an
initial value problem (1) exists in an x-interval larger than
that guaranteed by the present theorems. Show this fact
for y r ⫽ 2y 2, y(1) ⫽ 1 by finding the best possible a
9
(choosing b optimally) and comparing the result with the
actual solution.
6. CAS PROJECT. Picard Iteration. (a) Show that by
integrating the ODE in (1) and observing the initial
condition you obtain
x
(6)
y(x) ⫽ y0 ⫹
冮 f (t, y(t)) dt.
x0
This form (6) of (1) suggests Picard’s Iteration Method10
which is defined by
x
(7) yn(x) ⫽ y0 ⫹
冮 f (t, y
nⴚ1(t) dt,
n ⫽ 1, 2, Á .
x0
It gives approximations y1, y2, y3, . . . of the unknown
solution y of (1). Indeed, you obtain y1 by substituting
y ⫽ y0 on the right and integrating—this is the first
step—then y2 by substituting y ⫽ y1 on the right and
integrating—this is the second step—and so on. Write
RUDOLF LIPSCHITZ (1832–1903), German mathematician. Lipschitz and similar conditions are important
in modern theories, for instance, in partial differential equations.
10
EMILE PICARD (1856–1941). French mathematician, also known for his important contributions to
complex analysis (see Sec. 16.2 for his famous theorem). Picard used his method to prove Theorems 1 and 2
as well as the convergence of the sequence (7) to the solution of (1). In precomputer times, the iteration was of
little practical value because of the integrations.
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Chapter 1 Review Questions and Problems
a program of the iteration that gives a printout of the
first approximations y0, y1, . . . , yN as well as their
graphs on common axes. Try your program on two
initial value problems of your own choice.
(b) Apply the iteration to y r ⫽ x ⫹ y, y(0) ⫽ 0. Also
solve the problem exactly.
(c) Apply the iteration to y r ⫽ 2y 2, y(0) ⫽ 1. Also
solve the problem exactly.
(d) Find all solutions of y r ⫽ 2 1y, y(1) ⫽ 0. Which
of them does Picard’s iteration approximate?
(e) Experiment with the conjecture that Picard’s
iteration converges to the solution of the problem for
any initial choice of y in the integrand in (7) (leaving
y0 outside the integral as it is). Begin with a simple ODE
and see what happens. When you are reasonably sure,
take a slightly more complicated ODE and give it a try.
43
7. Maximum A. What is the largest possible a in
Example 1 in the text?
8. Lipschitz condition. Show that for a linear ODE
y r ⫹ p(x)y ⫽ r(x) with continuous p and r in
ƒ x ⫺ x 0 ƒ ⬉ a a Lipschitz condition holds. This is
remarkable because it means that for a linear ODE the
continuity of f (x, y) guarantees not only the existence
but also the uniqueness of the solution of an initial
value problem. (Of course, this also follows directly
from (4) in Sec. 1.5.)
9. Common points. Can two solution curves of the same
ODE have a common point in a rectangle in which the
assumptions of the present theorems are satisfied?
10. Three possible cases. Find all initial conditions such
that (x 2 ⫺ x)y r ⫽ (2x ⫺ 1)y has no solution, precisely
one solution, and more than one solution.
CHAPTER 1 REVIEW QUESTIONS AND PROBLEMS
1. Explain the basic concepts ordinary and partial
differential equations (ODEs, PDEs), order, general
and particular solutions, initial value problems (IVPs).
Give examples.
2. What is a linear ODE? Why is it easier to solve than
a nonlinear ODE?
3. Does every first-order ODE have a solution? A solution
formula? Give examples.
4. What is a direction field? A numeric method for firstorder ODEs?
5. What is an exact ODE? Is f (x) dx ⫹ g(y) dy ⫽ 0
always exact?
6. Explain the idea of an integrating factor. Give two
examples.
7. What other solution methods did we consider in this
chapter?
8. Can an ODE sometimes be solved by several methods?
Give three examples.
9. What does modeling mean? Can a CAS solve a model
given by a first-order ODE? Can a CAS set up a model?
10. Give problems from mechanics, heat conduction, and
population dynamics that can be modeled by first-order
ODEs.
11–16
14. xy r ⫽ y ⫹ x 2
15. y r ⫹ y ⫽ 1.01 cos 10x
16. Solve y r ⫽ y ⫺ y 2, y(0) ⫽ 0.2 by Euler’s method
(10 steps, h ⫽ 0.1). Solve exactly and compute the error.
17–21
GENERAL SOLUTION
Find the general solution. Indicate which method in this
chapter you are using. Show the details of your work.
17. y r ⫹ 2.5y ⫽ 1.6x
18. y r ⫺ 0.4y ⫽ 29 sin x
19. 25yy r ⫺ 4x ⫽ 0
20. y r ⫽ ay ⫹ by 2 (a ⫽ 0)
21. (3xey ⫹ 2y) dx ⫹ (x 2ey ⫹ x) dy ⫽ 0
22–26
INITIAL VALUE PROBLEM (IVP)
Solve the IVP. Indicate the method used. Show the details
of your work.
2
22. y r ⫹ 4xy ⫽ e⫺2x , y(0) ⫽ ⫺4.3
23. y r ⫽ 21 ⫺ y 2, y(0) ⫽ 1> 12
24. y r ⫹ 12 y ⫽ y 3, y(0) ⫽ 13
25. 3 sec y dx ⫹ 13 sec x dy ⫽ 0, y(0) ⫽ 0
26. x sinh y dy ⫽ cosh y dx, y(3) ⫽ 0
DIRECTION FIELD: NUMERIC SOLUTION
Graph a direction field (by a CAS or by hand) and sketch
some solution curves. Solve the ODE exactly and compare.
In Prob. 16 use Euler’s method.
11. y r ⫹ 2y ⫽ 0
12. y r ⫽ 1 ⫺ y 2
13. y r ⫽ y ⫺ 4y 2
27–30
MODELING, APPLICATIONS
27. Exponential growth. If the growth rate of a culture
of bacteria is proportional to the number of bacteria
present and after 1 day is 1.25 times the original
number, within what interval of time will the number
of bacteria (a) double, (b) triple?
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44
CHAP. 1 First-Order ODEs
28. Mixing problem. The tank in Fig. 28 contains 80 lb
of salt dissolved in 500 gal of water. The inflow per
minute is 20 lb of salt dissolved in 20 gal of water. The
outflow is 20 gal> min of the uniform mixture. Find the
time when the salt content y(t) in the tank reaches 95%
of its limiting value (as t : ⬁ ).
Fig. 28.
29. Half-life. If in a reactor, uranium 237
97 U loses 10% of
its weight within one day, what is its half-life? How
long would it take for 99% of the original amount to
disappear?
30. Newton’s law of cooling. A metal bar whose
temperature is 20°C is placed in boiling water. How
long does it take to heat the bar to practically 100°C,
say, to 99.9°C, if the temperature of the bar after 1 min
of heating is 51.5°C? First guess, then calculate.
Tank in Problem 28
SUMMARY OF CHAPTER
1
First-Order ODEs
This chapter concerns ordinary differential equations (ODEs) of first order and
their applications. These are equations of the form
(1)
F(x, y, y r ) ⫽ 0
or in explicit form
y r ⫽ f (x, y)
involving the derivative y r ⫽ dy>dx of an unknown function y, given functions of
x, and, perhaps, y itself. If the independent variable x is time, we denote it by t.
In Sec. 1.1 we explained the basic concepts and the process of modeling, that is,
of expressing a physical or other problem in some mathematical form and solving
it. Then we discussed the method of direction fields (Sec. 1.2), solution methods
and models (Secs. 1.3–1.6), and, finally, ideas on existence and uniqueness of
solutions (Sec. 1.7).
A first-order ODE usually has a general solution, that is, a solution involving an
arbitrary constant, which we denote by c. In applications we usually have to find a
unique solution by determining a value of c from an initial condition y(x 0) ⫽ y0.
Together with the ODE this is called an initial value problem
(2)
y r ⫽ f (x, y),
y(x 0) ⫽ y0
(x 0, y0 given numbers)
and its solution is a particular solution of the ODE. Geometrically, a general
solution represents a family of curves, which can be graphed by using direction
fields (Sec. 1.2). And each particular solution corresponds to one of these curves.
A separable ODE is one that we can put into the form
(3)
g(y) dy ⫽ f (x) dx
(Sec. 1.3)
by algebraic manipulations (possibly combined with transformations, such as
y>x ⫽ u) and solve by integrating on both sides.
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Summary of Chapter 1
45
An exact ODE is of the form
(4)
M(x, y) dx ⫹ N(x, y) dy ⫽ 0
(Sec. 1.4)
where M dx ⫹ N dy is the differential
du ⫽ u x dx ⫹ u y dy
of a function u(x, y), so that from du ⫽ 0 we immediately get the implicit general
solution u(x, y) ⫽ c. This method extends to nonexact ODEs that can be made exact
by multiplying them by some function F(x, y,), called an integrating factor (Sec. 1.4).
Linear ODEs
(5)
y r ⫹ p(x)y ⫽ r(x)
are very important. Their solutions are given by the integral formula (4), Sec. 1.5.
Certain nonlinear ODEs can be transformed to linear form in terms of new variables.
This holds for the Bernoulli equation
y r ⫹ p(x)y ⫽ g(x)y a
(Sec. 1.5).
Applications and modeling are discussed throughout the chapter, in particular in
Secs. 1.1, 1.3, 1.5 (population dynamics, etc.), and 1.6 (trajectories).
Picard’s existence and uniqueness theorems are explained in Sec. 1.7 (and
Picard’s iteration in Problem Set 1.7).
Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2
immediately after this chapter, as indicated in the chapter opening.
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CHAPTER
2
Second-Order Linear ODEs
Many important applications in mechanical and electrical engineering, as shown in Secs.
2.4, 2.8, and 2.9, are modeled by linear ordinary differential equations (linear ODEs) of the
second order. Their theory is representative of all linear ODEs as is seen when compared
to linear ODEs of third and higher order, respectively. However, the solution formulas for
second-order linear ODEs are simpler than those of higher order, so it is a natural progression
to study ODEs of second order first in this chapter and then of higher order in Chap. 3.
Although ordinary differential equations (ODEs) can be grouped into linear and nonlinear
ODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which many
beautiful standard methods exist.
Chapter 2 includes the derivation of general and particular solutions, the latter in
connection with initial value problems.
For those interested in solution methods for Legendre’s, Bessel’s, and the hypergeometric
equations consult Chap. 5 and for Sturm–Liouville problems Chap. 11.
COMMENT. Numerics for second-order ODEs can be studied immediately after this
chapter. See Sec. 21.3, which is independent of other sections in Chaps. 19–21.
Prerequisite: Chap. 1, in particular, Sec. 1.5.
Sections that may be omitted in a shorter course: 2.3, 2.9, 2.10.
References and Answers to Problems: App. 1 Part A, and App. 2.
2.1 Homogeneous Linear ODEs of Second Order
We have already considered first-order linear ODEs (Sec. 1.5) and shall now define and
discuss linear ODEs of second order. These equations have important engineering
applications, especially in connection with mechanical and electrical vibrations (Secs. 2.4,
2.8, 2.9) as well as in wave motion, heat conduction, and other parts of physics, as we
shall see in Chap. 12.
A second-order ODE is called linear if it can be written
(1)
y s p(x)y r q(x)y r(x)
and nonlinear if it cannot be written in this form.
The distinctive feature of this equation is that it is linear in y and its derivatives, whereas
the functions p, q, and r on the right may be any given functions of x. If the equation
begins with, say, f (x)y s, then divide by f (x) to have the standard form (1) with y s as the
first term.
46
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SEC. 2.1 Homogeneous Linear ODEs of Second Order
47
The definitions of homogeneous and nonhomogenous second-order linear ODEs are
very similar to those of first-order ODEs discussed in Sec. 1.5. Indeed, if r(x) ⬅ 0 (that
is, r(x) 0 for all x considered; read “r(x) is identically zero”), then (1) reduces to
y s p(x)y r q(x)y 0
(2)
and is called homogeneous. If r(x) [ 0, then (1) is called nonhomogeneous. This is
similar to Sec. 1.5.
An example of a nonhomogeneous linear ODE is
y s 25y eⴚx cos x,
and a homogeneous linear ODE is
xy s y r xy 0,
written in standard form
1
y s x y r y 0.
Finally, an example of a nonlinear ODE is
y s y y r 2 0.
The functions p and q in (1) and (2) are called the coefficients of the ODEs.
Solutions are defined similarly as for first-order ODEs in Chap. 1. A function
y h(x)
is called a solution of a (linear or nonlinear) second-order ODE on some open interval I
if h is defined and twice differentiable throughout that interval and is such that the ODE
becomes an identity if we replace the unknown y by h, the derivative y r by h r , and the
second derivative y s by h s . Examples are given below.
Homogeneous Linear ODEs: Superposition Principle
Sections 2.1–2.6 will be devoted to homogeneous linear ODEs (2) and the remaining
sections of the chapter to nonhomogeneous linear ODEs.
Linear ODEs have a rich solution structure. For the homogeneous equation the backbone
of this structure is the superposition principle or linearity principle, which says that we
can obtain further solutions from given ones by adding them or by multiplying them with
any constants. Of course, this is a great advantage of homogeneous linear ODEs. Let us
first discuss an example.
EXAMPLE 1
Homogeneous Linear ODEs: Superposition of Solutions
The functions y cos x and y sin x are solutions of the homogeneous linear ODE
ys y 0
for all x. We verify this by differentiation and substitution. We obtain (cos x) s cos x; hence
y s y (cos x) s cos x cos x cos x 0.
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CHAP. 2 Second-Order Linear ODEs
Similarly for y sin x (verify!). We can go an important step further. We multiply cos x by any constant, for
instance, 4.7, and sin x by, say, 2, and take the sum of the results, claiming that it is a solution. Indeed,
differentiation and substitution gives
(4.7 cos x 2 sin x) s (4.7 cos x 2 sin x) 4.7 cos x 2 sin x 4.7 cos x 2 sin x 0.
䊏
In this example we have obtained from y1 ( cos x) and y2 ( sin x) a function of the form
y c1y1 c2y2
(3)
(c1, c2 arbitrary constants).
This is called a linear combination of y1 and y2. In terms of this concept we can now
formulate the result suggested by our example, often called the superposition principle
or linearity principle.
THEOREM 1
Fundamental Theorem for the Homogeneous Linear ODE (2)
For a homogeneous linear ODE (2), any linear combination of two solutions on an
open interval I is again a solution of (2) on I. In particular, for such an equation,
sums and constant multiples of solutions are again solutions.
PROOF
Let y1 and y2 be solutions of (2) on I. Then by substituting y c1 y1 c2 y2 and
its derivatives into (2), and using the familiar rule (c1 y1 c2 y2) r c1 y1r c2 y 2r , etc.,
we get
y s py r qy (c1 y1 c2 y2) s p(c1 y1 c2 y2) r q(c1 y1 c2 y2)
c1 y1s c2 y s2 p(c1 y1r c2 y2r ) q(c1 y1 c2 y2)
c1( y1s py1r qy1) c2(y2s py 2r qy2) 0,
since in the last line, ( Á ) 0 because y1 and y2 are solutions, by assumption. This shows
that y is a solution of (2) on I.
䊏
CAUTION! Don’t forget that this highly important theorem holds for homogeneous
linear ODEs only but does not hold for nonhomogeneous linear or nonlinear ODEs, as
the following two examples illustrate.
EXAMPLE 2
A Nonhomogeneous Linear ODE
Verify by substitution that the functions y 1 cos x and y 1 sin x are solutions of the nonhomogeneous
linear ODE
y s y 1,
but their sum is not a solution. Neither is, for instance, 2(1 cos x) or 5(1 sin x).
EXAMPLE 3
䊏
A Nonlinear ODE
Verify by substitution that the functions y x 2 and y 1 are solutions of the nonlinear ODE
y s y xy r 0,
but their sum is not a solution. Neither is x 2, so you cannot even multiply by 1!
䊏
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SEC. 2.1 Homogeneous Linear ODEs of Second Order
49
Initial Value Problem. Basis. General Solution
Recall from Chap. 1 that for a first-order ODE, an initial value problem consists of the
ODE and one initial condition y(x 0) y0. The initial condition is used to determine the
arbitrary constant c in the general solution of the ODE. This results in a unique solution,
as we need it in most applications. That solution is called a particular solution of the
ODE. These ideas extend to second-order ODEs as follows.
For a second-order homogeneous linear ODE (2) an initial value problem consists of
(2) and two initial conditions
y(x 0) K 0,
(4)
y r (x 0) K 1.
These conditions prescribe given values K 0 and K 1 of the solution and its first derivative
(the slope of its curve) at the same given x x 0 in the open interval considered.
The conditions (4) are used to determine the two arbitrary constants c1 and c2 in a
general solution
y c1 y1 c2 y2
(5)
of the ODE; here, y1 and y2 are suitable solutions of the ODE, with “suitable” to be
explained after the next example. This results in a unique solution, passing through the
point (x 0, K 0) with K 1 as the tangent direction (the slope) at that point. That solution is
called a particular solution of the ODE (2).
EXAMPLE 4
Initial Value Problem
Solve the initial value problem
y s y 0,
Solution.
y(0) 3.0,
y r (0) 0.5.
Step 1. General solution. The functions cos x and sin x are solutions of the ODE (by Example 1),
and we take
y
y c1 cos x c2 sin x.
3
2
This will turn out to be a general solution as defined below.
1
Step 2. Particular solution. We need the derivative y r c1 sin x c2 cos x. From this and the
initial values we obtain, since cos 0 1 and sin 0 0,
0
2
4
6
8
10
–1
–2
x
y(0) c1 3.0
and
y r (0) c2 0.5.
This gives as the solution of our initial value problem the particular solution
–3
Fig. 29. Particular solution
and initial tangent in
Example 4
y 3.0 cos x 0.5 sin x.
Figure 29 shows that at x 0 it has the value 3.0 and the slope 0.5, so that its tangent intersects
䊏
the x-axis at x 3.0>0.5 6.0 . (The scales on the axes differ!)
Observation. Our choice of y1 and y2 was general enough to satisfy both initial
conditions. Now let us take instead two proportional solutions y1 cos x and y2 k cos x,
so that y1/y2 1/k const. Then we can write y c1 y1 c2 y2 in the form
y c1 cos x c2(k cos x) C cos x
where
C c1 c2k.
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CHAP. 2 Second-Order Linear ODEs
Hence we are no longer able to satisfy two initial conditions with only one arbitrary
constant C. Consequently, in defining the concept of a general solution, we must exclude
proportionality. And we see at the same time why the concept of a general solution is of
importance in connection with initial value problems.
DEFINITION
General Solution, Basis, Particular Solution
A general solution of an ODE (2) on an open interval I is a solution (5) in which
y1 and y2 are solutions of (2) on I that are not proportional, and c1 and c2 are arbitrary
constants. These y1, y2 are called a basis (or a fundamental system) of solutions
of (2) on I.
A particular solution of (2) on I is obtained if we assign specific values to c1
and c2 in (5).
For the definition of an interval see Sec. 1.1. Furthermore, as usual, y1 and y2 are called
proportional on I if for all x on I,
(6)
(a)
y1 ky2
or
(b)
y2 ly1
where k and l are numbers, zero or not. (Note that (a) implies (b) if and only if k 0).
Actually, we can reformulate our definition of a basis by using a concept of general
importance. Namely, two functions y1 and y2 are called linearly independent on an
interval I where they are defined if
(7)
k 1y1(x) k 2y2(x) 0
everywhere on I implies
k 1 0 and k 2 0.
And y1 and y2 are called linearly dependent on I if (7) also holds for some constants k 1,
k 2 not both zero. Then, if k 1 0 or k 2 0, we can divide and see that y1 and y2 are
proportional,
y1 k2
y2
k1
or
y2 k1
y1.
k2
In contrast, in the case of linear independence these functions are not proportional because
then we cannot divide in (7). This gives the following
DEFINITION
Basis (Reformulated)
A basis of solutions of (2) on an open interval I is a pair of linearly independent
solutions of (2) on I.
If the coefficients p and q of (2) are continuous on some open interval I, then (2) has a
general solution. It yields the unique solution of any initial value problem (2), (4). It
includes all solutions of (2) on I; hence (2) has no singular solutions (solutions not
obtainable from of a general solution; see also Problem Set 1.1). All this will be shown
in Sec. 2.6.
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SEC. 2.1 Homogeneous Linear ODEs of Second Order
EXAMPLE 5
51
Basis, General Solution, Particular Solution
cos x and sin x in Example 4 form a basis of solutions of the ODE y s y 0 for all x because their
quotient is cot x const (or tan x const). Hence y c1 cos x c2 sin x is a general solution. The solution
y 3.0 cos x 0.5 sin x of the initial value problem is a particular solution.
䊏
EXAMPLE 6
Basis, General Solution, Particular Solution
Verify by substitution that y1 ex and y2 eⴚx are solutions of the ODE y s y 0. Then solve the initial
value problem
y s y 0,
y(0) 6,
y r (0) 2.
Solution. (ex) s ex 0 and (eⴚx) s eⴚx 0 show that ex and eⴚx are solutions. They are not
proportional, ex/eⴚx e2x const. Hence ex, eⴚx form a basis for all x. We now write down the corresponding
general solution and its derivative and equate their values at 0 to the given initial conditions,
y c1ex c2eⴚx,
y r c1ex c2eⴚx,
y(0) c1 c2 6,
y r (0) c1 c2 2.
By addition and subtraction, c1 2, c2 4, so that the answer is y 2ex 4eⴚx. This is the particular solution
satisfying the two initial conditions.
䊏
Find a Basis if One Solution Is Known.
Reduction of Order
It happens quite often that one solution can be found by inspection or in some other way.
Then a second linearly independent solution can be obtained by solving a first-order ODE.
This is called the method of reduction of order.1 We first show how this method works
in an example and then in general.
EXAMPLE 7
Reduction of Order if a Solution Is Known. Basis
Find a basis of solutions of the ODE
(x 2 x)y s xy r y 0.
Inspection shows that y1 x is a solution because y1r 1 and y s1 0, so that the first term
vanishes identically and the second and third terms cancel. The idea of the method is to substitute
Solution.
y uy1 ux,
y r u r x u,
y s u s x 2u r
into the ODE. This gives
(x 2 x)(u s x 2u r ) x(u r x u) ux 0.
ux and –xu cancel and we are left with the following ODE, which we divide by x, order, and simplify,
(x 2 x)(u s x 2u r ) x 2u r 0,
(x 2 x)u s (x 2)u r 0.
This ODE is of first order in v u r , namely, (x 2 x)v r (x 2)v 0. Separation of variables and integration
gives
dv
1
2
x2
dx a
b dx,
2
x x
x1
v
x
1
ln ƒ v ƒ ln ƒ x 1 ƒ 2 ln ƒ x ƒ ln
ƒx 1ƒ
.
x2
Credited to the great mathematician JOSEPH LOUIS LAGRANGE (1736–1813), who was born in Turin,
of French extraction, got his first professorship when he was 19 (at the Military Academy of Turin), became
director of the mathematical section of the Berlin Academy in 1766, and moved to Paris in 1787. His important
major work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique analytique,
Paris, 1788), differential equations, approximation theory, algebra, and number theory.
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CHAP. 2 Second-Order Linear ODEs
We need no constant of integration because we want to obtain a particular solution; similarly in the next
integration. Taking exponents and integrating again, we obtain
v
x1
1
1
2,
x
x2
x
u
冮 v dx ln ƒ x ƒ x ,
1
y2 ux x ln ƒ x ƒ 1.
hence
Since y1 x and y2 x ln ƒ x ƒ 1 are linearly independent (their quotient is not constant), we have obtained
a basis of solutions, valid for all positive x.
䊏
In this example we applied reduction of order to a homogeneous linear ODE [see (2)]
y s p(x)y r q(x)y 0.
Note that we now take the ODE in standard form, with y s, not f (x)y s—this is essential
in applying our subsequent formulas. We assume a solution y1 of (2), on an open interval
I, to be known and want to find a basis. For this we need a second linearly independent
solution y2 of (2) on I. To get y2, we substitute
y y2 uy1,
y r y2r u r y1 uy1r ,
y s y2s u s y1 2u r y1r uy s1
into (2). This gives
(8)
u s y1 2u r y1r uy s1 p(u r y1 uy1r ) quy1 0.
Collecting terms in u s, u r, and u, we have
u s y1 u r (2y1r py1) u(y1s py 1r qy1) 0.
Now comes the main point. Since y1 is a solution of (2), the expression in the last
parentheses is zero. Hence u is gone, and we are left with an ODE in u r and u s . We divide
this remaining ODE by y1 and set u r U, u s U r,
us ur
2y1r py1
0,
y1
2y 1r
U r a y pb U 0.
thus
1
This is the desired first-order ODE, the reduced ODE. Separation of variables and
integration gives
2y1r
dU
a
pb dx
y1
U
and
ln ƒ U ƒ 2 ln ƒ y1 ƒ 冮 p dx.
By taking exponents we finally obtain
(9)
U
1 ⴚ兰p dx
e
.
y 21
Here U u r, so that u 兰 U dx. Hence the desired second solution is
冮
y2 y1u y1 U dx.
The quotient y2 /y1 u 兰 U dx cannot be constant (since U 0), so that y1 and y2 form
a basis of solutions.
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53
PROBLEM SET 2.1
REDUCTION OF ORDER is important because it
gives a simpler ODE. A general second-order ODE
F (x, y, y r , y s ) ⫽ 0, linear or not, can be reduced to first
order if y does not occur explicitly (Prob. 1) or if x does not
occur explicitly (Prob. 2) or if the ODE is homogeneous
linear and we know a solution (see the text).
1. Reduction. Show that F (x, y r, y s ) ⫽ 0 can be
reduced to first order in z ⫽ y r (from which y follows
by integration). Give two examples of your own.
2. Reduction. Show that F ( y, y r, y s ) ⫽ 0 can be
reduced to a first-order ODE with y as the independent
variable and y s ⫽ (dz/dy)z, where z ⫽ y r; derive this
by the chain rule. Give two examples.
3–10
REDUCTION OF ORDER
Reduce to first order and solve, showing each step in detail.
3. y s ⫹ y r ⫽ 0
4. 2xy s ⫽ 3y r
5. yy s ⫽ 3y r 2
6. xy s ⫹ 2y r ⫹ xy ⫽ 0, y1 ⫽ (cos x)/x
7. y s ⫹ y r 3 sin y ⫽ 0
8. y s ⫽ 1 ⫹ y r 2
9. x 2y s ⫺ 5xy r ⫹ 9y ⫽ 0, y1 ⫽ x 3
10. y s ⫹ (1 ⫹ 1/y)y r 2 ⫽ 0
11–14
APPLICATIONS OF REDUCIBLE ODEs
11. Curve. Find the curve through the origin in the
xy-plane which satisfies y s ⫽ 2y r and whose tangent
at the origin has slope 1.
12. Hanging cable. It can be shown that the curve y(x)
of an inextensible flexible homogeneous cable hanging
between two fixed points is obtained by solving
y s ⫽ k 21 ⫹ y r 2, where the constant k depends on the
weight. This curve is called catenary (from Latin
catena = the chain). Find and graph y(x), assuming that
k ⫽ 1 and those fixed points are (⫺1, 0) and (1, 0) in
a vertical xy-plane.
13. Motion. If, in the motion of a small body on a
straight line, the sum of velocity and acceleration equals
a positive constant, how will the distance y(t) depend
on the initial velocity and position?
14. Motion. In a straight-line motion, let the velocity be
the reciprocal of the acceleration. Find the distance y(t)
for arbitrary initial position and velocity.
15–19
GENERAL SOLUTION. INITIAL VALUE
PROBLEM (IVP)
(More in the next set.) (a) Verify that the given functions
are linearly independent and form a basis of solutions of
the given ODE. (b) Solve the IVP. Graph or sketch the
solution.
15. 4y s ⫹ 25y ⫽ 0, y(0) ⫽ 3.0, y r (0) ⫽ ⫺2.5,
cos 2.5x, sin 2.5x
16. y s ⫹ 0.6y r ⫹ 0.09y ⫽ 0, y(0) ⫽ 2.2, y r (0) ⫽ 0.14,
eⴚ0.3x, xeⴚ0.3x
17. 4x 2y s ⫺ 3y ⫽ 0, y(1) ⫽ ⫺3, y r (1) ⫽ 0,
x 3>2, x ⴚ1>2
18. x 2y s ⫺ xy r ⫹ y ⫽ 0, y(1) ⫽ 4.3, y r (1) ⫽ 0.5,
x, x ln x
19. y s ⫹ 2y r ⫹ 2y ⫽ 0, y(0) ⫽ 0, y r (0) ⫽ 15,
eⴚx cos x, eⴚx sin x
20. CAS PROJECT. Linear Independence. Write a
program for testing linear independence and dependence. Try it out on some of the problems in this and
the next problem set and on examples of your own.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
We shall now consider second-order homogeneous linear ODEs whose coefficients a and
b are constant,
(1)
y s ⫹ ay r ⫹ by ⫽ 0.
These equations have important applications in mechanical and electrical vibrations, as
we shall see in Secs. 2.4, 2.8, and 2.9.
To solve (1), we recall from Sec. 1.5 that the solution of the first-order linear ODE with
a constant coefficient k
y r ⫹ ky ⫽ 0
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CHAP. 2 Second-Order Linear ODEs
is an exponential function y ceⴚkx. This gives us the idea to try as a solution of (1) the
function
y elx.
(2)
Substituting (2) and its derivatives
y r lelx
and
y s l2elx
into our equation (1), we obtain
(l2 al b)elx 0.
Hence if l is a solution of the important characteristic equation (or auxiliary equation)
(3)
l2 al b 0
then the exponential function (2) is a solution of the ODE (1). Now from algebra we recall
that the roots of this quadratic equation (3) are
(4)
l1 12 Aa 2a 2 4b B ,
l2 12 Aa 2a 2 4b B .
(3) and (4) will be basic because our derivation shows that the functions
(5)
y1 el1x
and
y2 el2x
are solutions of (1). Verify this by substituting (5) into (1).
From algebra we further know that the quadratic equation (3) may have three kinds of
roots, depending on the sign of the discriminant a 2 4b, namely,
(Case I)
Two real roots if a 2 4b 0,
(Case II) A real double root if a 2 4b 0,
(Case III) Complex conjugate roots if a 2 4b 0.
Case I. Two Distinct Real-Roots l1 and l2
In this case, a basis of solutions of (1) on any interval is
y1 el1x
and
y2 el2x
because y1 and y2 are defined (and real) for all x and their quotient is not constant. The
corresponding general solution is
(6)
y c1el1x c2el2x.
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SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients
EXAMPLE 1
55
General Solution in the Case of Distinct Real Roots
We can now solve y s y 0 in Example 6 of Sec. 2.1 systematically. The characteristic equation is
l2 1 0. Its roots are l1 1 and l2 1. Hence a basis of solutions is ex and eⴚx and gives the same
general solution as before,
䊏
y c1ex c2eⴚx.
EXAMPLE 2
Initial Value Problem in the Case of Distinct Real Roots
Solve the initial value problem
y s y r 2y 0,
Solution.
y(0) 4,
y r (0) 5.
Step 1. General solution. The characteristic equation is
l2 l 2 0.
Its roots are
l1 12 (1 19 ) 1
and
l2 12 (1 19) 2
so that we obtain the general solution
y c1ex c2eⴚ2x.
Step 2. Particular solution. Since y r (x) c1ex 2c2eⴚ2x, we obtain from the general solution and the initial
conditions
y(0) c1 c2 4,
y r (0) c1 2c2 5.
Hence c1 1 and c2 3. This gives the answer y ex 3eⴚ2x. Figure 30 shows that the curve begins at
y 4 with a negative slope (5, but note that the axes have different scales!), in agreement with the initial
conditions.
䊏
y
8
6
4
2
0
0
0.5
1
1.5
2
x
Fig. 30. Solution in Example 2
Case II. Real Double Root l a/2
If the discriminant a 2 4b is zero, we see directly from (4) that we get only one root,
l l1 l2 a/2, hence only one solution,
y1 eⴚ(a/2)x.
To obtain a second independent solution y2 (needed for a basis), we use the method of
reduction of order discussed in the last section, setting y2 uy1. Substituting this and its
derivatives y r2 u r y1 uy 1r and y s2 into (1), we first have
(u sy1 2u r y 1r uy s1) a(u r y1 uy 1r ) buy1 0.
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CHAP. 2 Second-Order Linear ODEs
Collecting terms in u s, u r, and u, as in the last section, we obtain
u s y1 u r (2y 1r ay1) u(y s1 ay 1r by1) 0.
The expression in the last parentheses is zero, since y1 is a solution of (1). The expression
in the first parentheses is zero, too, since
2y 1r aeⴚax/2 ay1.
We are thus left with u s y1 0. Hence u s 0. By two integrations, u c1x c2. To
get a second independent solution y2 uy1, we can simply choose c1 1, c2 0 and
take u x. Then y2 xy1. Since these solutions are not proportional, they form a basis.
Hence in the case of a double root of (3) a basis of solutions of (1) on any interval is
eⴚax/2,
xeⴚax/2.
The corresponding general solution is
y (c1 c2x)eⴚax/2.
(7)
WARNING! If l is a simple root of (4), then (c1 c2x)elx with c2 0 is not a solution
of (1).
EXAMPLE 3
General Solution in the Case of a Double Root
The characteristic equation of the ODE y s 6y r 9y 0 is l2 6l 9 (l 3)2 0. It has the double
root l 3. Hence a basis is eⴚ3x and xeⴚ3x. The corresponding general solution is y (c1 c2x)eⴚ3x. 䊏
EXAMPLE 4
Initial Value Problem in the Case of a Double Root
Solve the initial value problem
y s y r 0.25y 0,
y(0) 3.0,
y r (0) 3.5.
The characteristic equation is l l 0.25 (l 0.5) 2 0. It has the double root l 0.5.
This gives the general solution
2
Solution.
y (c1 c2x)eⴚ0.5x.
We need its derivative
y r c2eⴚ0.5x 0.5(c1 c2x)eⴚ0.5x.
From this and the initial conditions we obtain
y(0) c1 3.0,
y r (0) c2 0.5c1 3.5;
The particular solution of the initial value problem is y (3 2x)e
c2 2.
hence
ⴚ0.5x
. See Fig. 31.
y
3
2
1
0
2
4
6
8
10
12
–1
Fig. 31. Solution in Example 4
14
x
䊏
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57
Case III. Complex Roots 21 a iv and 21 a iv
This case occurs if the discriminant a 2 4b of the characteristic equation (3) is negative.
In this case, the roots of (3) are the complex l 12 a iv that give the complex solutions
of the ODE (1). However, we will show that we can obtain a basis of real solutions
(8)
y1 eⴚax/2 cos vx,
y2 eⴚax/2 sin vx
(v 0)
where v2 b 14 a 2. It can be verified by substitution that these are solutions in the
present case. We shall derive them systematically after the two examples by using the
complex exponential function. They form a basis on any interval since their quotient
cot vx is not constant. Hence a real general solution in Case III is
y eⴚax/2 (A cos vx B sin vx)
(9)
EXAMPLE 5
(A, B arbitrary).
Complex Roots. Initial Value Problem
Solve the initial value problem
y s 0.4y r 9.04y 0,
y(0) 0,
y r (0) 3.
Step 1. General solution. The characteristic equation is l2 0.4l 9.04 0. It has the roots
0.2 3i. Hence v 3, and a general solution (9) is
Solution.
y eⴚ0.2x (A cos 3x B sin 3x).
Step 2. Particular solution. The first initial condition gives y(0) A 0. The remaining expression is
y Beⴚ0.2x sin 3x. We need the derivative (chain rule!)
y r B(0.2eⴚ0.2x sin 3x 3eⴚ0.2x cos 3x).
From this and the second initial condition we obtain y r (0) 3B 3. Hence B 1. Our solution is
y eⴚ0.2x sin 3x.
Figure 32 shows y and the curves of eⴚ0.2x and eⴚ0.2x (dashed), between which the curve of y oscillates.
Such “damped vibrations” (with x t being time) have important mechanical and electrical applications, as we
shall soon see (in Sec. 2.4).
䊏
y
1.0
0.5
0
5
10
15
20
25
30
x
–0.5
–1.0
Fig. 32.
EXAMPLE 6
Solution in Example 5
Complex Roots
A general solution of the ODE
y s v2y 0
(v constant, not zero)
is
y A cos vx B sin vx.
With v 1 this confirms Example 4 in Sec. 2.1.
䊏
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CHAP. 2 Second-Order Linear ODEs
Summary of Cases I–III
Case
Roots of (2)
Basis of (1)
General Solution of (1)
I
Distinct real
l1, l2
el1x, el2x
y c1el1x c2el2x
II
Real double root
l 12 a
eⴚax>2, xeⴚax>2
y (c1 c2x)eⴚax>2
III
Complex conjugate
l1 12 a iv,
l2 12 a iv
eⴚax>2 cos vx
y eⴚax>2(A cos vx B sin vx)
e
ⴚax>2
sin vx
It is very interesting that in applications to mechanical systems or electrical circuits,
these three cases correspond to three different forms of motion or flows of current,
respectively. We shall discuss this basic relation between theory and practice in detail in
Sec. 2.4 (and again in Sec. 2.8).
Derivation in Case III. Complex Exponential Function
If verification of the solutions in (8) satisfies you, skip the systematic derivation of these
real solutions from the complex solutions by means of the complex exponential function
ez of a complex variable z r it. We write r it, not x iy because x and y occur
in the ODE. The definition of ez in terms of the real functions er, cos t, and sin t is
(10)
ez erit ereit er(cos t i sin t).
This is motivated as follows. For real z r, hence t 0, cos 0 1, sin 0 0, we get
the real exponential function er. It can be shown that ez1z2 ez1ez2, just as in real. (Proof
in Sec. 13.5.) Finally, if we use the Maclaurin series of ez with z it as well as
i 2 1, i 3 i, i 4 1, etc., and reorder the terms as shown (this is permissible, as
can be proved), we obtain the series
eit 1 it 1
(it)2
(it)3
(it)4
(it) 5 Á
2!
3!
4!
5!
t2
t4
t3
t5
Á i at Áb
2!
4!
3!
5!
cos t i sin t.
(Look up these real series in your calculus book if necessary.) We see that we have obtained
the formula
(11)
eit cos t i sin t,
called the Euler formula. Multiplication by er gives (10).
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SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients
59
For later use we note that eⴚit cos (t) i sin (t) cos t i sin t, so that by
addition and subtraction of this and (11),
cos t 12 (eit eⴚit),
(12)
sin t 1 it
(e eⴚit).
2i
After these comments on the definition (10), let us now turn to Case III.
In Case III the radicand a 2 4b in (4) is negative. Hence 4b a 2 is positive and,
using 11 i, we obtain in (4)
1
1
1 2
1 2
2
2
2 2a 4b 2 2(4b a ) 2(b 4 a ) i 2b 4 a iv
with v defined as in (8). Hence in (4),
l1 12 a iv
and, similarly,
l2 12 a iv.
Using (10) with r 12 ax and t vx, we thus obtain
el1x eⴚ(a/2)xivx eⴚ(a/2)x(cos vx i sin vx)
el2x eⴚ(a/2)xivx eⴚ(a/2)x(cos vx i sin vx).
We now add these two lines and multiply the result by 12. This gives y1 as in (8). Then
we subtract the second line from the first and multiply the result by 1/(2i). This gives y2
as in (8). These results obtained by addition and multiplication by constants are again
solutions, as follows from the superposition principle in Sec. 2.1. This concludes the
derivation of these real solutions in Case III.
PROBLEM SET 2.2
1–15
GENERAL SOLUTION
Find a general solution. Check your answer by substitution.
ODEs of this kind have important applications to be
discussed in Secs. 2.4, 2.7, and 2.9.
1. 4y s 25y 0
2. y s 36y 0
3. y s 6y r 8.96y 0
4. y s 4y r (p2 4)y 0
5. y s 2py r p2y 0
6. 10y s 32y r 25.6y 0
7. y s 4.5y r 0
8. y s y r 3.25y 0
9. y s 1.8y r 2.08y 0
10. 100y s 240y r (196p2 144)y 0
11. 4y s 4y r 3y 0
12. y s 9y r 20y 0
13. 9y s 30y r 25y 0
14. y s 2k 2y r k 4y 0
15. y s 0.54y r (0.0729 p)y 0
16–20
FIND AN ODE
y s ay r by 0 for the given basis.
16. e2.6x, eⴚ4.3x
17. eⴚ25x, xeⴚ25x
18. cos 2px, sin 2px
19. e(ⴚ2i)x, e(ⴚ2ⴚi)x
ⴚ3.1x
ⴚ3.1x
20. e
cos 2.1x, e
sin 2.1x
21–30
INITIAL VALUES PROBLEMS
Solve the IVP. Check that your answer satisfies the ODE as
well as the initial conditions. Show the details of your work.
21. y s 25y 0, y(0) 4.6, y r (0) 1.2
22. The ODE in Prob. 4, y(12) 1, y r (12) 2
23. y s y r 6y 0, y(0) 10, y r (0) 0
24. 4y s 4y r 3y 0, y(2) e, y r (2) e>2
25. y s y 0, y(0) 2, y r (0) 2
26. y s k 2y 0 (k 0), y(0) 1, y r (0) 1
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CHAP. 2 Second-Order Linear ODEs
27. The ODE in Prob. 5,
y(0) 4.5, y r (0) 4.5p 1 13.137
28. 8y s 2y r y 0, y(0) 0.2, y r (0) 0.325
29. The ODE in Prob. 15, y(0) 0, y r (0) 1
30. 9y s 30y r 25y 0, y(0) 3.3, y r (0) 10.0
31–36
LINEAR INDEPENDENCE is of basic importance, in this chapter, in connection with general solutions,
as explained in the text. Are the following functions linearly
independent on the given interval? Show the details of your
work.
31. ekx, xekx, any interval
32. eax, eⴚax, x 0
33. x 2, x 2 ln x, x 1
34. ln x, ln (x 3), x 1
35. sin 2x, cos x sin x, x 0
36. eⴚx cos 12 x, 0, 1 x 1
37. Instability. Solve y s y 0 for the initial conditions
y(0) 1, y r (0) 1. Then change the initial conditions
to y(0) 1.001, y r (0) 0.999 and explain why this
small change of 0.001 at t 0 causes a large change later,
e.g., 22 at t 10. This is instability: a small initial
difference in setting a quantity (a current, for instance) becomes larger and larger with time t. This is
undesirable.
38. TEAM PROJECT. General Properties of Solutions
(a) Coefficient formulas. Show how a and b in (1)
can be expressed in terms of l1 and l2. Explain how
these formulas can be used in constructing equations
for given bases.
(b) Root zero. Solve y s 4y r 0 (i) by the present
method, and (ii) by reduction to first order. Can you
explain why the result must be the same in both
cases? Can you do the same for a general ODE
y s ay r 0?
(c) Double root. Verify directly that xelx with l a>2 is a solution of (1) in the case of a double root.
Verify and explain why y eⴚ2x is a solution of
y s y r 6y 0 but xe2x is not.
(d) Limits. Double roots should be limiting cases of
distinct roots l1, l2 as, say, l2 : l1. Experiment with
this idea. (Remember l’Hôpital’s rule from calculus.)
Can you arrive at xel1x? Give it a try.
2.3 Differential Operators.
Optional
This short section can be omitted without interrupting the flow of ideas. It will not be
used subsequently, except for the notations Dy, D 2 y, etc. to stand for y r , y s , etc.
Operational calculus means the technique and application of operators. Here, an
operator is a transformation that transforms a function into another function. Hence
differential calculus involves an operator, the differential operator D, which
transforms a (differentiable) function into its derivative. In operator notation we write
d
D dx
and
(1)
Dy y r dy
.
dx
Similarly, for the higher derivatives we write D 2y D(Dy) y s , and so on. For example,
D sin cos, D 2 sin sin, etc.
For a homogeneous linear ODE y s ay r by 0 with constant coefficients we can
now introduce the second-order differential operator
L P(D) D 2 aD bI,
where I is the identity operator defined by Iy y. Then we can write that ODE as
(2)
Ly P(D)y (D 2 aD bI)y 0.
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SEC. 2.3 Differential Operators. Optional
61
P suggests “polynomial.” L is a linear operator. By definition this means that if Ly and
Lw exist (this is the case if y and w are twice differentiable), then L(cy kw) exists for
any constants c and k, and
L(cy kw) cLy kLw.
Let us show that from (2) we reach agreement with the results in Sec. 2.2. Since
(Del)(x) lelx and (D 2el)(x) l2elx, we obtain
Lel(x) P(D)el(x) (D 2 aD bI)el(x)
(3)
(l2 al b)elx P(l)elx 0.
This confirms our result of Sec. 2.2 that elx is a solution of the ODE (2) if and only if l
is a solution of the characteristic equation P(l) 0.
P(l) is a polynomial in the usual sense of algebra. If we replace l by the operator D,
we obtain the “operator polynomial” P(D). The point of this operational calculus is that
P(D) can be treated just like an algebraic quantity. In particular, we can factor it.
EXAMPLE 1
Factorization, Solution of an ODE
Factor P(D) D 2 3D 40I and solve P(D)y 0.
D 2 3D 40I (D 8I )(D 5I ) because I 2 I. Now (D 8I)y y r 8y 0 has the
solution y1 e8x. Similarly, the solution of (D 5I )y 0 is y2 eⴚ5x. This is a basis of P(D)y 0 on any
interval. From the factorization we obtain the ODE, as expected,
Solution.
(D 8I )(D 5I )y (D 8I )(y r 5y) D(y r 5y) 8(y r 5y)
y s 5y r 8y r 40y y s 3 r 40y 0.
Verify that this agrees with the result of our method in Sec. 2.2. This is not unexpected because we factored
䊏
P(D) in the same way as the characteristic polynomial P(l) l2 3l 40.
It was essential that L in (2) had constant coefficients. Extension of operator methods to
variable-coefficient ODEs is more difficult and will not be considered here.
If operational methods were limited to the simple situations illustrated in this section,
it would perhaps not be worth mentioning. Actually, the power of the operator approach
appears in more complicated engineering problems, as we shall see in Chap. 6.
PROBLEM SET 2.3
1–5
APPLICATION OF DIFFERENTIAL
OPERATORS
Apply the given operator to the given functions. Show all
steps in detail.
1. D 2 2D; cosh 2x, eⴚx e2x, cos x
2. D 3I; 3x 2 3x, 3e3x, cos 4x sin 4x
3. (D 2I )2; e2x, xe2x, eⴚ2x
4. (D 6I )2; 6x sin 6x, xeⴚ6x
5. (D 2I )(D 3I );
e2x, xe2x, eⴚ3x
6–12
GENERAL SOLUTION
Factor as in the text and solve.
6. (D 2 4.00D 3.36I )y 0
7. (4D 2 I )y 0
8. (D 2 3I )y 0
9. (D 2 4.20D 4.41I )y 0
10. (D 2 4.80D 5.76I )y 0
11. (D 2 4.00D 3.84I )y 0
12. (D 2 3.0D 2.5I )y 0
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CHAP. 2 Second-Order Linear ODEs
13. Linear operator. Illustrate the linearity of L in (2) by
taking c 4, k 6, y e2x, and w cos 2x.
Prove that L is linear.
14. Double root. If D 2 aD bI has distinct roots
␮ and l, show that a particular solution is
y (e␮x elx)>(␮ l). Obtain from this a solution
xelx by letting ␮ : l and applying l’Hôpital’s rule.
15. Definition of linearity. Show that the definition of
linearity in the text is equivalent to the following. If
L[ y] and L[w] exist, then L[ y w] exists and L[cy]
and L[kw] exist for all constants c and k, and
L[ y w] L[ y] L[w] as well as L[cy] cL[ y]
and L[kw] kL[w].
2.4 Modeling of Free Oscillations
of a Mass–Spring System
Linear ODEs with constant coefficients have important applications in mechanics, as we
show in this section as well as in Sec. 2.8, and in electrical circuits as we show in Sec. 2.9.
In this section we model and solve a basic mechanical system consisting of a mass on an
elastic spring (a so-called “mass–spring system,” Fig. 33), which moves up and down.
Setting Up the Model
We take an ordinary coil spring that resists extension as well as compression. We suspend
it vertically from a fixed support and attach a body at its lower end, for instance, an iron
ball, as shown in Fig. 33. We let y 0 denote the position of the ball when the system
is at rest (Fig. 33b). Furthermore, we choose the downward direction as positive, thus
regarding downward forces as positive and upward forces as negative.
Unstretched
spring
s0
(y = 0)
y
System at
rest
(a)
Fig. 33.
(b)
System in
motion
(c)
Mechanical mass–spring system
We now let the ball move, as follows. We pull it down by an amount y 0 (Fig. 33c).
This causes a spring force
(1)
F1 ky
(Hooke’s law2)
proportional to the stretch y, with k ( 0) called the spring constant. The minus sign
indicates that F1 points upward, against the displacement. It is a restoring force: It wants
to restore the system, that is, to pull it back to y 0. Stiff springs have large k.
2
ROBERT HOOKE (1635–1703), English physicist, a forerunner of Newton with respect to the law of
gravitation.
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SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System
63
Note that an additional force F0 is present in the spring, caused by stretching it in
fastening the ball, but F0 has no effect on the motion because it is in equilibrium with
the weight W of the ball, F0 W mg, where g 980 cm>sec2 9.8 m>sec2 32.17 ft>sec2 is the constant of gravity at the Earth’s surface (not to be confused with
the universal gravitational constant G gR2>M 6.67 # 10ⴚ11 nt m2>kg 2, which we
shall not need; here R 6.37 # 106 m and M 5.98 # 1024 kg are the Earth’s radius and
mass, respectively).
The motion of our mass–spring system is determined by Newton’s second law
(2)
Acceleration my s Force
Mass
where y s d 2y>dt 2 and “Force” is the resultant of all the forces acting on the ball. (For
systems of units, see the inside of the front cover.)
ODE of the Undamped System
Every system has damping. Otherwise it would keep moving forever. But if the damping
is small and the motion of the system is considered over a relatively short time, we
may disregard damping. Then Newton’s law with F F1 gives the model
my s F1 ky; thus
my s ky 0.
(3)
This is a homogeneous linear ODE with constant coefficients. A general solution is
obtained as in Sec. 2.2, namely (see Example 6 in Sec. 2.2)
y(t) A cos v0t B sin v0t
(4)
v0 k
.
m
B
This motion is called a harmonic oscillation (Fig. 34). Its frequency is f v0>2p Hertz3
( cycles>sec) because cos and sin in (4) have the period 2p>v0. The frequency f is called
the natural frequency of the system. (We write v0 to reserve v for Sec. 2.8.)
y
2
1
t
3
1 Positive
2 Zero
3 Negative
Initial velocity
Fig. 34. Typical harmonic oscillations (4) and (4*) with the same y(0) A and
different initial velocities y r (0) v0 B, positive 1 , zero 2 , negative 3
3
HEINRICH HERTZ (1857–1894), German physicist, who discovered electromagnetic waves, as the basis
of wireless communication developed by GUGLIELMO MARCONI (1874–1937), Italian physicist (Nobel prize
in 1909).
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CHAP. 2 Second-Order Linear ODEs
An alternative representation of (4), which shows the physical characteristics of amplitude
and phase shift of (4), is
y(t) C cos (v0t d)
(4*)
with C 2A2 B 2 and phase angle d, where tan d B>A. This follows from the
addition formula (6) in App. 3.1.
EXAMPLE 1
Harmonic Oscillation of an Undamped Mass–Spring System
If a mass–spring system with an iron ball of weight W 98 nt (about 22 lb) can be regarded as undamped, and
the spring is such that the ball stretches it 1.09 m (about 43 in.), how many cycles per minute will the system
execute? What will its motion be if we pull the ball down from rest by 16 cm (about 6 in.) and let it start with
zero initial velocity?
Hooke’s law (1) with W as the force and 1.09 meter as the stretch gives W 1.09k; thus
k W>1.09 98>1.09 90 [kg>sec2] 90 [nt>meter]. The mass is m W>g 98>9.8 10 [kg]. This
gives the frequency v0>(2p) 2k>m>(2p) 3>(2p) 0.48 [Hz] 29 [cycles>min].
From (4) and the initial conditions, y(0) A 0.16 [meter] and y r (0) v0B 0. Hence the motion is
Solution.
y(t) 0.16 cos 3t [meter]
or
0.52 cos 3t [ft]
(Fig. 35).
If you have a chance of experimenting with a mass–spring system, don’t miss it. You will be surprised about
the good agreement between theory and experiment, usually within a fraction of one percent if you measure
䊏
carefully.
y
0.2
0.1
0
2
–0.1
–0.2
Fig. 35.
4
6
8
10
t
Harmonic oscillation in Example 1
ODE of the Damped System
To our model my s ky we now add a damping force
F2 cy r ,
k
Spring
m
Ball
obtaining my s ky cy r ; thus the ODE of the damped mass–spring system is
(5)
c
Dashpot
Fig. 36.
Damped system
my s cy r ky 0.
(Fig. 36)
Physically this can be done by connecting the ball to a dashpot; see Fig. 36. We assume
this damping force to be proportional to the velocity y r dy>dt. This is generally a good
approximation for small velocities.
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SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System
65
The constant c is called the damping constant. Let us show that c is positive. Indeed,
the damping force F2 cy r acts against the motion; hence for a downward motion we
have y r 0 which for positive c makes F negative (an upward force), as it should be.
Similarly, for an upward motion we have y r 0 which, for c 0 makes F2 positive (a
downward force).
The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solve
it by the method in Sec. 2.2. The characteristic equation is (divide (5) by m)
c
k
l2 m l m 0.
By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2,
(6) l1 a b, l2 a b, where
a
c
2m
and
b
1
2c2 4mk.
2m
It is now interesting that depending on the amount of damping present—whether a lot of
damping, a medium amount of damping or little damping—three types of motions occur,
respectively:
Case I.
c2 4mk.
Distinct real roots l1, l2.
(Overdamping)
Case II.
c2 4mk.
A real double root.
(Critical damping)
Complex conjugate roots.
(Underdamping)
Case III. c2 4mk .
They correspond to the three Cases I, II, III in Sec. 2.2.
Discussion of the Three Cases
Case I. Overdamping
If the damping constant c is so large that c2 4mk, then l1 and l2 are distinct real roots.
In this case the corresponding general solution of (5) is
(7)
y(t) c1eⴚ(aⴚb)t c2eⴚ(aⴙb)t.
We see that in this case, damping takes out energy so quickly that the body does not
oscillate. For t 0 both exponents in (7) are negative because a 0, b 0, and
b2 a2 k>m a2. Hence both terms in (7) approach zero as t : . Practically
speaking, after a sufficiently long time the mass will be at rest at the static equilibrium
position (y 0). Figure 37 shows (7) for some typical initial conditions.
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CHAP. 2 Second-Order Linear ODEs
y
y
1
t
1
2
2
3
3
t
(a)
(b)
1 Positive
2 Zero
3 Negative
Initial velocity
Fig. 37. Typical motions (7) in the overdamped case
(a) Positive initial displacement
(b) Negative initial displacement
Case II. Critical Damping
Critical damping is the border case between nonoscillatory motions (Case I) and oscillations
(Case III). It occurs if the characteristic equation has a double root, that is, if c2 4mk,
so that b 0, l1 l2 a. Then the corresponding general solution of (5) is
y(t) (c1 c2t)eⴚat.
(8)
This solution can pass through the equilibrium position y 0 at most once because eⴚat
is never zero and c1 c2t can have at most one positive zero. If both c1 and c2 are positive
(or both negative), it has no positive zero, so that y does not pass through 0 at all. Figure 38
shows typical forms of (8). Note that they look almost like those in the previous figure.
y
1
2
3
t
1 Positive
2 Zero
3 Negative
Fig. 38.
Initial velocity
Critical damping [see (8)]
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SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System
67
Case III. Underdamping
This is the most interesting case. It occurs if the damping constant c is so small that
c2 4mk. Then b in (6) is no longer real but pure imaginary, say,
(9)
b iv*
where
v* c2
1
k
24mk c2 4m 2
2m
Bm
(0).
(We now write v* to reserve v for driving and electromotive forces in Secs. 2.8 and 2.9.)
The roots of the characteristic equation are now complex conjugates,
l1 a iv*,
l2 a iv*
with a c>(2m), as given in (6). Hence the corresponding general solution is
(10)
y(t) eⴚat(A cos v*t B sin v*t) Ceⴚat cos (v*t d)
where C 2 A2 B 2 and tan d B>A, as in (4*).
This represents damped oscillations. Their curve lies between the dashed curves
y Ceⴚat and y Ceⴚat in Fig. 39, touching them when v*t d is an integer multiple
of p because these are the points at which cos (v*t d) equals 1 or 1.
The frequency is v*>(2p) Hz (hertz, cycles/sec). From (9) we see that the smaller
c (0) is, the larger is v* and the more rapid the oscillations become. If c approaches 0,
then v* approaches v0 2k>m, giving the harmonic oscillation (4), whose frequency
v0>(2p) is the natural frequency of the system.
y
–α t
Ce
t
–α t
–Ce
Fig. 39.
EXAMPLE 2
Damped oscillation in Case III [see (10)]
The Three Cases of Damped Motion
How does the motion in Example 1 change if we change the damping constant c from one to another of the
following three values, with y(0) 0.16 and y r (0) 0 as before?
(I) c 100 kg>sec,
(II) c 60 kg>sec,
(III) c 10 kg>sec.
Solution.
It is interesting to see how the behavior of the system changes due to the effect of the damping,
which takes energy from the system, so that the oscillations decrease in amplitude (Case III) or even disappear
(Cases II and I).
(I) With m 10 and k 90, as in Example 1, the model is the initial value problem
10y s 100y r 90y 0,
y(0) 0.16 [meter],
y r (0) 0.
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CHAP. 2 Second-Order Linear ODEs
The characteristic equation is 10l2 100l 90 10(l 9)(l 1) 0. It has the roots 9 and 1. This
gives the general solution
y c1eⴚ9t c2eⴚt.
We also need
y r 9c1eⴚ9t c2eⴚt.
The initial conditions give c1 c2 0.16, 9c1 c2 0. The solution is c1 0.02, c2 0.18. Hence in
the overdamped case the solution is
y 0.02eⴚ9t 0.18eⴚt.
It approaches 0 as t : . The approach is rapid; after a few seconds the solution is practically 0, that is, the
iron ball is at rest.
(II) The model is as before, with c 60 instead of 100. The characteristic equation now has the form
10l2 60l 90 10(l 3) 2 0. It has the double root 3. Hence the corresponding general solution is
y (c1 c2t)eⴚ3t.
We also need
y r (c2 3c1 3c2t)eⴚ3t.
The initial conditions give y(0) c1 0.16, y r (0) c2 3c1 0, c2 0.48. Hence in the critical case the
solution is
y (0.16 0.48t)eⴚ3t.
It is always positive and decreases to 0 in a monotone fashion.
(III) The model now is 10y s 10y r 90y 0. Since c 10 is smaller than the critical c, we shall get
oscillations. The characteristic equation is 10l2 10l 90 10[(l 12 ) 2 9 14 ] 0. It has the complex
roots [see (4) in Sec. 2.2 with a 1 and b 9]
l 0.5 20.52 9 0.5 2.96i.
This gives the general solution
y eⴚ0.5t(A cos 2.96t B sin 2.96t).
Thus y(0) A 0.16. We also need the derivative
y r eⴚ0.5t(0.5A cos 2.96t 0.5B sin 2.96t 2.96A sin 2.96t 2.96B cos 2.96t).
Hence y r (0) 0.5A 2.96B 0, B 0.5A>2.96 0.027. This gives the solution
y eⴚ0.5t(0.16 cos 2.96t 0.027 sin 2.96t) 0.162eⴚ0.5t cos (2.96t 0.17).
We see that these damped oscillations have a smaller frequency than the harmonic oscillations in Example 1 by
about 1% (since 2.96 is smaller than 3.00 by about 1% ). Their amplitude goes to zero. See Fig. 40.
䊏
y
0.15
0.1
0.05
0
2
4
6
8
10
t
–0.05
–0.1
Fig. 40. The three solutions in Example 2
This section concerned free motions of mass–spring systems. Their models are homogeneous linear ODEs. Nonhomogeneous linear ODEs will arise as models of forced
motions, that is, motions under the influence of a “driving force.” We shall study them
in Sec. 2.8, after we have learned how to solve those ODEs.
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SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System
69
PROBLEM SET 2.4
1–10
HARMONIC OSCILLATIONS
(UNDAMPED MOTION)
1. Initial value problem. Find the harmonic motion (4)
that starts from y0 with initial velocity v0. Graph or
sketch the solutions for v0 p, y0 1, and various
v0 of your choice on common axes. At what t-values
do all these curves intersect? Why?
2. Frequency. If a weight of 20 nt (about 4.5 lb) stretches
a certain spring by 2 cm, what will the frequency of the
corresponding harmonic oscillation be? The period?
3. Frequency. How does the frequency of the harmonic
oscillation change if we (i) double the mass, (ii) take
a spring of twice the modulus? First find qualitative
answers by physics, then look at formulas.
4. Initial velocity. Could you make a harmonic oscillation
move faster by giving the body a greater initial push?
5. Springs in parallel. What are the frequencies of
vibration of a body of mass m 5 kg (i) on a spring
of modulus k 1 20 nt>m, (ii) on a spring of modulus
k 2 45 nt>m, (iii) on the two springs in parallel? See
Fig. 41.
The cylindrical buoy of diameter 60 cm in Fig. 43 is
floating in water with its axis vertical. When depressed
downward in the water and released, it vibrates with
period 2 sec. What is its weight?
Water
level
Fig. 43. Buoy (Problem 8)
9. Vibration of water in a tube. If 1 liter of water (about
1.06 US quart) is vibrating up and down under the
influence of gravitation in a U-shaped tube of diameter
2 cm (Fig. 44), what is the frequency? Neglect friction.
First guess.
y
( y = 0)
Fig. 44. Tube (Problem 9)
Fig. 41. Parallel springs (Problem 5)
6. Spring in series. If a body hangs on a spring s1 of
modulus k 1 8, which in turn hangs on a spring s2
of modulus k 2 12, what is the modulus k of this
combination of springs?
7. Pendulum. Find the frequency of oscillation of a
pendulum of length L (Fig. 42), neglecting air
resistance and the weight of the rod, and assuming u
to be so small that sin u practically equals u.
L
θ
Body of
mass m
10. TEAM PROJECT. Harmonic Motions of Similar
Models. The unifying power of mathematical methods results to a large extent from the fact that different
physical (or other) systems may have the same or very
similar models. Illustrate this for the following three
systems
(a) Pendulum clock. A clock has a 1-meter pendulum.
The clock ticks once for each time the pendulum
completes a full swing, returning to its original position.
How many times a minute does the clock tick?
(b) Flat spring (Fig. 45). The harmonic oscillations
of a flat spring with a body attached at one end and
horizontally clamped at the other are also governed by
(3). Find its motions, assuming that the body weighs
8 nt (about 1.8 lb), the system has its static equilibrium
1 cm below the horizontal line, and we let it start from
this position with initial velocity 10 cm/sec.
Fig. 42. Pendulum (Problem 7)
8. Archimedian principle. This principle states that the
buoyancy force equals the weight of the water
displaced by the body (partly or totally submerged).
y
Fig. 45. Flat spring
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CHAP. 2 Second-Order Linear ODEs
(c) Torsional vibrations (Fig. 46). Undamped
torsional vibrations (rotations back and forth) of a
wheel attached to an elastic thin rod or wire are
governed by the equation I0u s Ku 0, where u
is the angle measured from the state of equilibrium.
Solve this equation for K>I0 13.69 secⴚ2, initial
angle 30°( 0.5235 rad) and initial angular velocity
20° secⴚ1 ( 0.349 rad # secⴚ1).
θ
Fig. 46. Torsional vibrations
11–20
DAMPED MOTION
11. Overdamping. Show that for (7) to satisfy initial conditions y(0) y0 and v(0) v0 we must have c1 [(1 a>b)y0 v0>b]>2 and c2 [(1 a>b)y0 v0>b]>2.
12. Overdamping. Show that in the overdamped case, the
body can pass through y 0 at most once (Fig. 37).
13. Initial value problem. Find the critical motion (8)
that starts from y0 with initial velocity v0. Graph
solution curves for a 1, y0 1 and several v0 such
that (i) the curve does not intersect the t-axis, (ii) it
intersects it at t 1, 2, . . . , 5, respectively.
14. Shock absorber. What is the smallest value of the
damping constant of a shock absorber in the suspension of a wheel of a car (consisting of a spring and an
absorber) that will provide (theoretically) an oscillationfree ride if the mass of the car is 2000 kg and the spring
constant equals 4500 kg>sec 2?
15. Frequency. Find an approximation formula for v* in
terms of v0 by applying the binomial theorem in (9)
and retaining only the first two terms. How good is the
approximation in Example 2, III?
16. Maxima. Show that the maxima of an underdamped
motion occur at equidistant t-values and find the
distance.
equals ¢ 2pa>v*. Find ¢ for the solutions of
y s 2y r 5y 0.
19. Damping constant. Consider an underdamped motion
of a body of mass m 0.5 kg. If the time between two
consecutive maxima is 3 sec and the maximum
amplitude decreases to 12 its initial value after 10 cycles,
what is the damping constant of the system?
20. CAS PROJECT. Transition Between Cases I, II,
III. Study this transition in terms of graphs of typical
solutions. (Cf. Fig. 47.)
(a) Avoiding unnecessary generality is part of good
modeling. Show that the initial value problems (A)
and (B),
(A) y s cy r y 0,
y(0) 1,
y r (0) 0
(B) the same with different c and y r (0) 2 (instead
of 0), will give practically as much information as a
problem with other m, k, y(0), y r (0).
(b) Consider (A). Choose suitable values of c,
perhaps better ones than in Fig. 47, for the transition
from Case III to II and I. Guess c for the curves in the
figure.
(c) Time to go to rest. Theoretically, this time is
infinite (why?). Practically, the system is at rest when
its motion has become very small, say, less than 0.1%
of the initial displacement (this choice being up to us),
that is in our case,
(11)
ƒ y(t) ƒ 0.001
for all t greater than some t 1.
In engineering constructions, damping can often be
varied without too much trouble. Experimenting with
your graphs, find empirically a relation between t 1
and c.
(d) Solve (A) analytically. Give a reason why the
solution c of y(t 2) 0.001, with t 2 the solution of
y r (t) 0, will give you the best possible c satisfying
(11).
(e) Consider (B) empirically as in (a) and (b). What
is the main difference between (B) and (A)?
y
1
17. Underdamping. Determine the values of t corresponding to the maxima and minima of the oscillation
y(t) eⴚt sin t. Check your result by graphing y(t).
0.5
18. Logarithmic decrement. Show that the ratio of
two consecutive maximum amplitudes of a damped
oscillation (10) is constant, and the natural logarithm
of this ratio called the logarithmic decrement,
– 0.5
2
4
6
8
–1
Fig. 47. CAS Project 20
10
t
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SEC. 2.5 Euler–Cauchy Equations
71
2.5 Euler–Cauchy Equations
Euler–Cauchy equations4 are ODEs of the form
x 2y s axy r by 0
(1)
with given constants a and b and unknown function y(x). We substitute
y x m,
y r mx mⴚ1,
y s m(m 1)x mⴚ2
into (1). This gives
x 2m(m 1)x m2 axmx m1 bx m 0
and we now see that y x m was a rather natural choice because we have obtained a common factor x m. Dropping it, we have the auxiliary equation m(m 1) am b 0 or
(2)
m 2 (a 1)m b 0.
(Note: a 1, not a.)
Hence y x m is a solution of (1) if and only if m is a root of (2). The roots of (2) are
(3) m 1 12 (1 a) 214 (1 a)2 b,
m 2 12 (1 a) 214 (1 a)2 b.
Case I. Real different roots m 1 and m 2 give two real solutions
y1(x) x m1
y2(x) x m2.
and
These are linearly independent since their quotient is not constant. Hence they constitute
a basis of solutions of (1) for all x for which they are real. The corresponding general
solution for all these x is
(4)
EXAMPLE 1
y c1x m1 c2x m2
(c1, c2 arbitrary).
General Solution in the Case of Different Real Roots
The Euler–Cauchy equation x 2y s 1.5xy r 0.5y 0 has the auxiliary equation m 2 0.5m 0.5 0. The
roots are 0.5 and 1. Hence a basis of solutions for all positive x is y1 x 0.5 and y2 1>x and gives the general
solution
y c1 1x 4
c2
x
(x 0).
䊏
LEONHARD EULER (1707–1783) was an enormously creative Swiss mathematician. He made
fundamental contributions to almost all branches of mathematics and its application to physics. His important
books on algebra and calculus contain numerous basic results of his own research. The great French
mathematician AUGUSTIN LOUIS CAUCHY (1789–1857) is the father of modern analysis. He is the creator
of complex analysis and had great influence on ODEs, PDEs, infinite series, elasticity theory, and optics.
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72
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CHAP. 2 Second-Order Linear ODEs
Case II. A real double root m 1 12 (1 a) occurs if and only if b 14 (a 1)2 because
then (2) becomes [m 12 (a 1)]2, as can be readily verified. Then a solution is
y1 x (1ⴚa)>2, and (1) is of the form
(1 a)
a
y 0.
yr x
4x 2
2
(5)
x 2y s axy r 14 (1 a)2y 0
ys or
A second linearly independent solution can be obtained by the method of reduction of
order from Sec. 2.1, as follows. Starting from y2 uy1, we obtain for u the expression
(9) Sec. 2.1, namely,
冮
u U dx
U
where
冮
1
exp a p dxb .
y 12
From (5) in standard form (second ODE) we see that p a>x (not ax; this is essential!).
Hence exp 兰 (p dx) exp (a ln x) exp (ln x ⴚa) 1>x a. Division by y 12 x 1 a
gives U 1>x, so that u ln x by integration. Thus, y2 uy1 y1 ln x, and y1 and y2
are linearly independent since their quotient is not constant. The general solution
corresponding to this basis is
(6)
EXAMPLE 2
y (c1 c2 ln x) x m,
m 12 (1 a).
General Solution in the Case of a Double Root
The Euler–Cauchy equation x 2y s 5xy r 9y 0 has the auxiliary equation m 2 6m 9 0. It has the
double root m 3, so that a general solution for all positive x is
y (c1 c2 ln x) x 3.
䊏
Case III. Complex conjugate roots are of minor practical importance, and we discuss
the derivation of real solutions from complex ones just in terms of a typical example.
EXAMPLE 3
Real General Solution in the Case of Complex Roots
The Euler–Cauchy equation x 2y s 0.6xy r 16.04y 0 has the auxiliary equation m 2 0.4m 16.04 0.
The roots are complex conjugate, m 1 0.2 4i and m 2 0.2 4i, where i 11. We now use the trick
of writing x eln x and obtain
x m1 x 0.24i x 0.2(eln x)4i x 0.2e(4 ln x)i,
x m2 x 0.2ⴚ4i x 0.2(eln x)ⴚ4i x 0.2eⴚ(4 ln x)i.
Next we apply Euler’s formula (11) in Sec. 2.2 with t 4 ln x to these two formulas. This gives
x m1 x 0.2[cos (4 ln x) i sin (4 ln x)],
x m2 x 0.2[cos (4 ln x) i sin (4 ln x)].
We now add these two formulas, so that the sine drops out, and divide the result by 2. Then we subtract the
second formula from the first, so that the cosine drops out, and divide the result by 2i. This yields
x 0.2 cos (4 ln x)
and
x 0.2 sin (4 ln x)
respectively. By the superposition principle in Sec. 2.2 these are solutions of the Euler–Cauchy equation (1).
Since their quotient cot (4 ln x) is not constant, they are linearly independent. Hence they form a basis of solutions,
and the corresponding real general solution for all positive x is
(8)
y x 0.2[A cos (4 ln x) B sin (4 ln x)].
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SEC. 2.5 Euler–Cauchy Equations
73
Figure 48 shows typical solution curves in the three cases discussed, in particular the real basis functions in
Examples 1 and 3.
䊏
y
y
x 1.5
3.0
x1
2.0
x 0.5
1.0
x –0.5
x –1.5
0
1
2
x
Case I: Real roots
1.5
1.0
0.5
x 0.5 ln x
0
–0.5
–1.0
–1.5
x –1
y
x ln x
1.5
1.0
0.5
0.4
x –0.5 ln x
x –1.5 ln x
2 x
1 1.4
0
–0.5
–1.0
–1.5
Case II: Double root
x 0.2 sin (4 ln x)
0.4
1 1.4
2
x
x 0.2 cos (4 ln x)
Case III: Complex roots
Fig. 48. Euler–Cauchy equations
EXAMPLE 4
Boundary Value Problem. Electric Potential Field Between Two Concentric Spheres
Find the electrostatic potential v v(r) between two concentric spheres of radii r1 5 cm and r2 10 cm
kept at potentials v1 110 V and v2 0, respectively.
Physical Information. v(r) is a solution of the Euler–Cauchy equation rv s 2v r 0, where v r dv>dr.
Solution. The auxiliary equation is m2 m 0. It has the roots 0 and 1. This gives the general solution
v(r) c1 c2>r. From the “boundary conditions” (the potentials on the spheres) we obtain
v(5) c1 c2
110.
5
v(10) c1 c2
0.
10
By subtraction, c2>10 110, c2 1100. From the second equation, c1 c2>10 110. Answer:
v(r) 110 1100>r V. Figure 49 shows that the potential is not a straight line, as it would be for a potential
between two parallel plates. For example, on the sphere of radius 7.5 cm it is not 110>2 55 V, but considerably
䊏
less. (What is it?)
v
100
80
60
40
20
0
5
6
7
8
9
10
r
Fig. 49. Potential v(r) in Example 4
PROBLEM SET 2.5
1. Double root. Verify directly by substitution that
x (1ⴚa)>2 ln x is a solution of (1) if (2) has a double root,
but x m1 ln x and x m2 ln x are not solutions of (1) if the
roots m1 and m2 of (2) are different.
2–11
GENERAL SOLUTION
Find a real general solution. Show the details of your work.
2. x 2y s 20y 0
3. 5x 2y s 23xy r 16.2y 0
4. xy s 2y r 0
5. 4x 2y s 5y 0
6. x 2y s 0.7xy r 0.1y 0
7. (x 2D 2 4xD 6I)y C
8. (x 2D 2 3xD 4I)y 0
9. (x 2D 2 0.2xD 0.36I)y 0
10. (x 2D 2 xD 5I)y 0
11. (x 2D 2 3xD 10I)y 0
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12–19
Page 74
CHAP. 2 Second-Order Linear ODEs
INITIAL VALUE PROBLEM
Solve and graph the solution. Show the details of your work.
12. x 2y s 4xy r 6y 0, y(1) 0.4, y r (1) 0
13. x 2y s 3xy r 0.75y 0, y(1) 1,
y r (1) 1.5
14. x 2y s xy r 9y 0, y(1) 0, y r (1) 2.5
15. x 2y s 3xy r y 0, y(1) 3.6, y r (1) 0.4
16. (x 2D 2 3xD 4I )y 0, y(1) p, y r (1) 2p
17. (x 2D 2 xD I )y 0, y(1) 1, y r (1) 1
18. (9x 2D 2 3xD I )y 0, y(1) 1, y r (1) 0
19. (x 2D 2 xD 15I )y 0, y(1) 0.1,
y r (1) 4.5
20. TEAM PROJECT. Double Root
(a) Derive a second linearly independent solution of
(1) by reduction of order; but instead of using (9), Sec.
2.1, perform all steps directly for the present ODE (1).
(b) Obtain x m ln x by considering the solutions x m
and x ms of a suitable Euler–Cauchy equation and
letting s : 0.
(c) Verify by substitution that x m ln x, m (1 a)>2,
is a solution in the critical case.
(d) Transform the Euler–Cauchy equation (1) into
an ODE with constant coefficients by setting
x et (x 0).
(e) Obtain a second linearly independent solution of
the Euler–Cauchy equation in the “critical case” from
that of a constant-coefficient ODE.
2.6 Existence and Uniqueness
of Solutions. Wronskian
In this section we shall discuss the general theory of homogeneous linear ODEs
(1)
y s p(x)y r q(x)y 0
with continuous, but otherwise arbitrary, variable coefficients p and q. This will concern
the existence and form of a general solution of (1) as well as the uniqueness of the solution
of initial value problems consisting of such an ODE and two initial conditions
(2)
y(x 0) K 0,
y r (x 0) K 1
with given x 0, K 0, and K 1.
The two main results will be Theorem 1, stating that such an initial value problem
always has a solution which is unique, and Theorem 4, stating that a general solution
(3)
y c1y1 c2y2
(c1, c2 arbitrary)
includes all solutions. Hence linear ODEs with continuous coefficients have no “singular
solutions” (solutions not obtainable from a general solution).
Clearly, no such theory was needed for constant-coefficient or Euler–Cauchy equations
because everything resulted explicitly from our calculations.
Central to our present discussion is the following theorem.
THEOREM 1
Existence and Uniqueness Theorem for Initial Value Problems
If p(x) and q(x) are continuous functions on some open interval I (see Sec. 1.1) and
x0 is in I, then the initial value problem consisting of (1) and (2) has a unique
solution y(x) on the interval I.
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SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian
75
The proof of existence uses the same prerequisites as the existence proof in Sec. 1.7
and will not be presented here; it can be found in Ref. [A11] listed in App. 1. Uniqueness
proofs are usually simpler than existence proofs. But for Theorem 1, even the uniqueness
proof is long, and we give it as an additional proof in App. 4.
Linear Independence of Solutions
Remember from Sec. 2.1 that a general solution on an open interval I is made up from a
basis y1, y2 on I, that is, from a pair of linearly independent solutions on I. Here we call
y1, y2 linearly independent on I if the equation
k1y1(x) k 2y2(x) 0
(4)
on I
implies
k1 0, k 2 0.
We call y1, y2 linearly dependent on I if this equation also holds for constants k 1, k 2
not both 0. In this case, and only in this case, y1 and y2 are proportional on I, that is (see
Sec. 2.1),
(a) y1 ky2
(5)
or
(b) y2 ly1
for all on I.
For our discussion the following criterion of linear independence and dependence of
solutions will be helpful.
THEOREM 2
Linear Dependence and Independence of Solutions
Let the ODE (1) have continuous coefficients p(x) and q(x) on an open interval I.
Then two solutions y1 and y2 of (1) on I are linearly dependent on I if and only if
their “Wronskian”
(6)
W(y1, y2) y1y2r y2y1r
is 0 at some x 0 in I. Furthermore, if W 0 at an x x 0 in I, then W 0 on I;
hence, if there is an x 1 in I at which W is not 0, then y1, y2 are linearly independent
on I.
PROOF
(a) Let y1 and y2 be linearly dependent on I. Then (5a) or (5b) holds on I. If (5a) holds,
then
W(y1, y2) y1y2r y2y1r ky2y2r y2ky2r 0.
Similarly if (5b) holds.
(b) Conversely, we let W( y1, y2) 0 for some x x 0 and show that this implies linear
dependence of y1 and y2 on I. We consider the linear system of equations in the unknowns
k 1, k 2
(7)
k 1 y1(x 0) k 2 y2(x 0) 0
k 1 y1r (x 0) k 2 y2r (x 0) 0.
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CHAP. 2 Second-Order Linear ODEs
To eliminate k 2, multiply the first equation by y 2r and the second by y2 and add the
resulting equations. This gives
k 1y1(x 0)y2r (x 0) k 1y1r (x 0)y2(x 0) k 1W( y1(x 0), y2(x 0)) 0.
Similarly, to eliminate k 1, multiply the first equation by y1r and the second by y1 and
add the resulting equations. This gives
k 2W( y1(x 0), y2(x 0)) 0.
If W were not 0 at x 0, we could divide by W and conclude that k 1 k 2 0. Since W is
0, division is not possible, and the system has a solution for which k 1 and k 2 are not both
0. Using these numbers k 1, k 2, we introduce the function
y(x) k 1y1(x) k 2y2(x).
Since (1) is homogeneous linear, Fundamental Theorem 1 in Sec. 2.1 (the superposition
principle) implies that this function is a solution of (1) on I. From (7) we see that it satisfies
the initial conditions y(x 0) 0, y r (x 0) 0. Now another solution of (1) satisfying the
same initial conditions is y* ⬅ 0. Since the coefficients p and q of (1) are continuous,
Theorem 1 applies and gives uniqueness, that is, y ⬅ y*, written out
k 1y1 k 2y2 ⬅ 0
on I.
Now since k 1 and k 2 are not both zero, this means linear dependence of y1, y2 on I.
(c) We prove the last statement of the theorem. If W(x 0) 0 at an x 0 in I, we have
linear dependence of y1, y2 on I by part (b), hence W ⬅ 0 by part (a) of this proof. Hence
in the case of linear dependence it cannot happen that W(x 1) 0 at an x 1 in I. If it does
happen, it thus implies linear independence as claimed.
䊏
For calculations, the following formulas are often simpler than (6).
(6*) W( y1, y2) (a)
y2 r
a y b y 12
1
( y1 0)
or
(b)
y1 r
a y b y 22
2
( y2 0).
These formulas follow from the quotient rule of differentiation.
Remark. Determinants. Students familiar with second-order determinants may have
noticed that
W( y1, y2) `
y1
y2
y1r
y 2r
` y1y 2r y2y1r .
This determinant is called the Wronski determinant5 or, briefly, the Wronskian, of two
solutions y1 and y2 of (1), as has already been mentioned in (6). Note that its four entries
occupy the same positions as in the linear system (7).
5
Introduced by WRONSKI (JOSEF MARIA HÖNE, 1776–1853), Polish mathematician.
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SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian
EXAMPLE 1
77
Illustration of Theorem 2
The functions y1 cos vx and y2 sin vx are solutions of y s v2y 0. Their Wronskian is
W(cos vx, sin vx) `
cos vx
sin vx
v sin vx
v cos vx
` y1y2r y2y1r v cos2 vx v sin2 vx v.
Theorem 2 shows that these solutions are linearly independent if and only if v 0. Of course, we can see
this directly from the quotient y2>y1 tan vx. For v 0 we have y2 0, which implies linear dependence
䊏
(why?).
EXAMPLE 2
Illustration of Theorem 2 for a Double Root
A general solution of y s 2y r y 0 on any interval is y (c1 c2x)ex. (Verify!). The corresponding
Wronskian is not 0, which shows linear independence of ex and xex on any interval. Namely,
ex
W(x, xex) ` x
e
xex
(x 1)ex
` (x 1)e2x xe2x e2x 0.
䊏
A General Solution of (1) Includes All Solutions
This will be our second main result, as announced at the beginning. Let us start with
existence.
THEOREM 3
Existence of a General Solution
If p(x) and q(x) are continuous on an open interval I, then (1) has a general solution
on I.
PROOF
By Theorem 1, the ODE (1) has a solution y1(x) on I satisfying the initial conditions
y1(x 0) 1,
y1r (x 0) 0
and a solution y2(x) on I satisfying the initial conditions
y2(x 0) 0,
y2r (x 0) 1.
The Wronskian of these two solutions has at x x 0 the value
W( y1(0), y2(0)) y1(x 0)y2r (x 0) y2(x 0)y1r (x 0) 1.
Hence, by Theorem 2, these solutions are linearly independent on I. They form a basis of
solutions of (1) on I, and y c1y1 c2y2 with arbitrary c1, c2 is a general solution of (1)
on I, whose existence we wanted to prove.
䊏
˛
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CHAP. 2 Second-Order Linear ODEs
We finally show that a general solution is as general as it can possibly be.
THEOREM 4
A General Solution Includes All Solutions
If the ODE (1) has continuous coefficients p(x) and q(x) on some open interval I,
then every solution y Y(x) of (1) on I is of the form
Y(x) C1y1(x) C2y2(x)
(8)
where y1, y2 is any basis of solutions of (1) on I and C1, C2 are suitable constants.
Hence (1) does not have singular solutions (that is, solutions not obtainable from
a general solution).
PROOF
Let y Y(x) be any solution of (1) on I. Now, by Theorem 3 the ODE (1) has a general
solution
y(x) c1y1(x) c2y2(x)
(9)
on I. We have to find suitable values of c1, c2 such that y(x) Y(x) on I. We choose any
x 0 in I and show first that we can find values of c1, c2 such that we reach agreement at
x 0, that is, y(x 0) Y(x 0) and y r (x 0) Y r (x 0). Written out in terms of (9), this becomes
(10)
(a)
c1y1(x 0) c2y2(x 0) Y(x 0)
(b) c1y1r (x 0) c2y2r (x 0) Y r (x 0).
We determine the unknowns c1 and c2. To eliminate c2, we multiply (10a) by y2r (x 0) and
(10b) by y2(x 0) and add the resulting equations. This gives an equation for c1. Then we
multiply (10a) by y1r (x 0) and (10b) by y1(x 0) and add the resulting equations. This gives
an equation for c2. These new equations are as follows, where we take the values of
y1, y1r , y2, y2r , Y, Y r at x 0.
c1( y1y2r y2y1r ) c1W( y1, y2) Yy2r y2Y r
c2( y1y2r y2y1r ) c2W( y1, y2) y1Y r Yy1r .
Since y1, y2 is a basis, the Wronskian W in these equations is not 0, and we can solve for
c1 and c2. We call the (unique) solution c1 C1, c2 C2. By substituting it into (9) we
obtain from (9) the particular solution
y*(x) C1y1(x) C2 y2(x).
Now since C1, C2 is a solution of (10), we see from (10) that
y*(x 0) Y(x 0),
y* r (x 0) Y r (x 0).
From the uniqueness stated in Theorem 1 this implies that y* and Y must be equal
everywhere on I, and the proof is complete.
䊏
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SEC. 2.7 Nonhomogeneous ODEs
79
Reflecting on this section, we note that homogeneous linear ODEs with continuous variable
coefficients have a conceptually and structurally rather transparent existence and uniqueness
theory of solutions. Important in itself, this theory will also provide the foundation for our
study of nonhomogeneous linear ODEs, whose theory and engineering applications form
the content of the remaining four sections of this chapter.
PROBLEM SET 2.6
1. Derive (6*) from (6).
2–8
BASIS OF SOLUTIONS. WRONSKIAN
Find the Wronskian. Show linear independence by using
quotients and confirm it by Theorem 2.
2. e4.0x, eⴚ1.5x
3. eⴚ0.4x, eⴚ2.6x
4. x, 1>x
5. x 3, x 2
6. eⴚx cos vx, eⴚx sin vx
7. cosh ax, sinh ax
8. x k cos (ln x), x k sin (ln x)
9–15
ODE FOR GIVEN BASIS. WRONSKIAN. IVP
(a) Find a second-order homogeneous linear ODE for
which the given functions are solutions. (b) Show linear
independence by the Wronskian. (c) Solve the initial value
problem.
9. cos 5x, sin 5x, y(0) 3, y r (0) 5
10. x m1, x m2, y(1) 2, y r (1) 2m 1 4m 2
11. eⴚ2.5x cos 0.3x, eⴚ2.5x sin 0.3x, y(0) 3,
y r (0) 7.5
12. x 2, x 2 ln x, y(1) 4, y r (1) 6
13. 1, e2x, y(0) 1, y r (0) 1
14. ekx cos px, ekx sin px, y(0) 1,
y r (0) k p
15. cosh 1.8x, sinh 1.8x, y(0) 14.20, y r (0) 16.38
16. TEAM PROJECT. Consequences of the Present
Theory. This concerns some noteworthy general
properties of solutions. Assume that the coefficients p
and q of the ODE (1) are continuous on some open
interval I, to which the subsequent statements refer.
(a) Solve y s y 0 (a) by exponential functions,
(b) by hyperbolic functions. How are the constants in
the corresponding general solutions related?
(b) Prove that the solutions of a basis cannot be 0 at
the same point.
(c) Prove that the solutions of a basis cannot have a
maximum or minimum at the same point.
(d) Why is it likely that formulas of the form (6*)
should exist?
(e) Sketch y1(x) x 3 if x
0 and 0 if x 0,
y2(x) 0 if x
0 and x 3 if x 0. Show linear
independence on 1 x 1. What is their
Wronskian? What Euler–Cauchy equation do y1, y2
satisfy? Is there a contradiction to Theorem 2?
(f) Prove Abel’s formula6
W( y1(x), y2(x)) c exp c x
冮 p(t) dt d
x0
where c W(y1(x 0), y2(x 0)). Apply it to Prob. 6. Hint:
Write (1) for y1 and for y2. Eliminate q algebraically
from these two ODEs, obtaining a first-order linear
ODE. Solve it.
2.7 Nonhomogeneous ODEs
We now advance from homogeneous to nonhomogeneous linear ODEs.
Consider the second-order nonhomogeneous linear ODE
(1)
y s p(x)y r q(x)y r(x)
where r(x) [ 0. We shall see that a “general solution” of (1) is the sum of a general
solution of the corresponding homogeneous ODE
6
NIELS HENRIK ABEL (1802–1829), Norwegian mathematician.
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CHAP. 2 Second-Order Linear ODEs
y s p(x)y r q(x)y 0
(2)
and a “particular solution” of (1). These two new terms “general solution of (1)” and
“particular solution of (1)” are defined as follows.
DEFINITION
General Solution, Particular Solution
A general solution of the nonhomogeneous ODE (1) on an open interval I is a
solution of the form
(3)
y(x) yh(x) yp1x2;
here, yh c1y1 c2y2 is a general solution of the homogeneous ODE (2) on I and
yp is any solution of (1) on I containing no arbitrary constants.
A particular solution of (1) on I is a solution obtained from (3) by assigning
specific values to the arbitrary constants c1 and c2 in yh.
Our task is now twofold, first to justify these definitions and then to develop a method
for finding a solution yp of (1).
Accordingly, we first show that a general solution as just defined satisfies (1) and that
the solutions of (1) and (2) are related in a very simple way.
THEOREM 1
Relations of Solutions of (1) to Those of (2)
(a) The sum of a solution y of (1) on some open interval I and a solution ~y of
(2) on I is a solution of (1) on I. In particular, (3) is a solution of (1) on I.
(b) The difference of two solutions of (1) on I is a solution of (2) on I.
PROOF
(a) Let L[y] denote the left side of (1). Then for any solutions y of (1) and ~y of (2) on I,
L[ y ~y ] L[ y] L[ ~y ] r 0 r.
(b) For any solutions y and y* of (1) on I we have L[ y y*] L[ y] L[ y*] r r 0.
䊏
Now for homogeneous ODEs (2) we know that general solutions include all solutions.
We show that the same is true for nonhomogeneous ODEs (1).
THEOREM 2
A General Solution of a Nonhomogeneous ODE Includes All Solutions
If the coefficients p(x), q(x), and the function r(x) in (1) are continuous on some
open interval I, then every solution of (1) on I is obtained by assigning suitable
values to the arbitrary constants c1 and c2 in a general solution (3) of (1) on I.
PROOF
Let y* be any solution of (1) on I and x 0 any x in I. Let (3) be any general solution of
(1) on I. This solution exists. Indeed, yh c1y1 c2y2 exists by Theorem 3 in Sec. 2.6
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SEC. 2.7 Nonhomogeneous ODEs
81
because of the continuity assumption, and yp exists according to a construction to be
shown in Sec. 2.10. Now, by Theorem 1(b) just proved, the difference Y y* yp is a
solution of (2) on I. At x 0 we have
Y1x 02 y*1x 02 yp(x 0).
Y r 1x 02 y* r 1x 02 ypr 1x 02.
Theorem 1 in Sec. 2.6 implies that for these conditions, as for any other initial conditions
in I, there exists a unique particular solution of (2) obtained by assigning suitable values
䊏
to c1, c2 in yh. From this and y* Y yp the statement follows.
Method of Undetermined Coefficients
Our discussion suggests the following. To solve the nonhomogeneous ODE (1) or an initial
value problem for (1), we have to solve the homogeneous ODE (2) and find any solution
yp of (1), so that we obtain a general solution (3) of (1).
How can we find a solution yp of (1)? One method is the so-called method of
undetermined coefficients. It is much simpler than another, more general, method (given
in Sec. 2.10). Since it applies to models of vibrational systems and electric circuits to be
shown in the next two sections, it is frequently used in engineering.
More precisely, the method of undetermined coefficients is suitable for linear ODEs
with constant coefficients a and b
(4)
y s ay r by r(x)
when r (x) is an exponential function, a power of x, a cosine or sine, or sums or products
of such functions. These functions have derivatives similar to r (x) itself. This gives the
idea. We choose a form for yp similar to r (x), but with unknown coefficients to be
determined by substituting that yp and its derivatives into the ODE. Table 2.1 on p. 82
shows the choice of yp for practically important forms of r (x). Corresponding rules are
as follows.
Choice Rules for the Method of Undetermined Coefficients
(a) Basic Rule. If r (x) in (4) is one of the functions in the first column in
Table 2.1, choose yp in the same line and determine its undetermined
coefficients by substituting yp and its derivatives into (4).
(b) Modification Rule. If a term in your choice for yp happens to be a
solution of the homogeneous ODE corresponding to (4), multiply this term
by x (or by x 2 if this solution corresponds to a double root of the
characteristic equation of the homogeneous ODE).
(c) Sum Rule. If r (x) is a sum of functions in the first column of Table 2.1,
choose for yp the sum of the functions in the corresponding lines of the
second column.
The Basic Rule applies when r (x) is a single term. The Modification Rule helps in the
indicated case, and to recognize such a case, we have to solve the homogeneous ODE
first. The Sum Rule follows by noting that the sum of two solutions of (1) with r r1
and r r2 (and the same left side!) is a solution of (1) with r r1 r2. (Verify!)
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CHAP. 2 Second-Order Linear ODEs
The method is self-correcting. A false choice for yp or one with too few terms will lead
to a contradiction. A choice with too many terms will give a correct result, with superfluous
coefficients coming out zero.
Let us illustrate Rules (a)–(c) by the typical Examples 1–3.
Table 2.1
EXAMPLE 1
Method of Undetermined Coefficients
Term in r (x)
Choice for yp(x)
kegx
kx n (n 0, 1, Á )
k cos vx
k sin vx
keax cos vx
keax sin vx
Cegx
K nx n K n1x n1 Á K 1x K 0
f K cos vx M sin vx
f eax(K cos vx M sin vx)
Application of the Basic Rule (a)
Solve the initial value problem
y s y 0.001x 2,
(5)
Solution.
y(0) 0,
y r (0) 1.5.
Step 1. General solution of the homogeneous ODE. The ODE y s y 0 has the general solution
yh A cos x B sin x.
Step 2. Solution yp of the nonhomogeneous ODE. We first try yp Kx 2. Then y sp 2K. By substitution,
2K Kx 2 0.001x 2. For this to hold for all x, the coefficient of each power of x (x 2 and x 0) must be the same
on both sides; thus K 0.001 and 2K 0, a contradiction.
The second line in Table 2.1 suggests the choice
yp K 2 x 2 K 1x K 0.
Then
y sp yp 2K 2 K 2x 2 K 1x K 0 0.001x 2.
Equating the coefficients of x 2, x, x 0 on both sides, we have K 2 0.001, K 1 0, 2K 2 K 0 0. Hence
K 0 2K 2 0.002. This gives yp 0.001x 2 0.002, and
y yh yp A cos x B sin x 0.001x 2 0.002.
Step 3. Solution of the initial value problem. Setting x 0 and using the first initial condition gives
y(0) A 0.002 0, hence A 0.002. By differentiation and from the second initial condition,
y r yhr ypr A sin x B cos x 0.002x
y r (0) B 1.5.
and
This gives the answer (Fig. 50)
y 0.002 cos x 1.5 sin x 0.001x 2 0.002.
Figure 50 shows y as well as the quadratic parabola yp about which y is oscillating, practically like a sine curve
since the cosine term is smaller by a factor of about 1>1000.
䊏
y
2
1
0
–1
10
Fig. 50.
20
30
40
Solution in Example 1
x
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SEC. 2.7 Nonhomogeneous ODEs
EXAMPLE 2
83
Application of the Modification Rule (b)
Solve the initial value problem
y s 3y r 2.25y 10eⴚ1.5x,
(6)
y(0) 1,
y r (0) 0.
Solution.
Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous
ODE is l2 3l 2.25 (l 1.5)2 0. Hence the homogeneous ODE has the general solution
yh (c1 c2x)eⴚ1.5x.
˛
Step 2. Solution yp of the nonhomogeneous ODE. The function eⴚ1.5x on the right would normally require
the choice Ceⴚ1.5x. But we see from yh that this function is a solution of the homogeneous ODE, which
corresponds to a double root of the characteristic equation. Hence, according to the Modification Rule we have
to multiply our choice function by x 2. That is, we choose
yp Cx 2eⴚ1.5x.
Then
ypr C(2x 1.5x 2)eⴚ1.5x,
y sp C(2 3x 3x 2.25x 2)eⴚ1.5x.
We substitute these expressions into the given ODE and omit the factor eⴚ1.5x. This yields
C(2 6x 2.25x 2) 3C(2x 1.5x 2) 2.25Cx 2 10.
Comparing the coefficients of x 2, x, x 0 gives 0 0, 0 0, 2C 10, hence C 5. This gives the solution
yp 5x 2eⴚ1.5x. Hence the given ODE has the general solution
y yh yp (c1 c2x)eⴚ1.5x 5x 2eⴚ1.5x.
Step 3. Solution of the initial value problem. Setting x 0 in y and using the first initial condition, we obtain
y(0) c1 1. Differentiation of y gives
y r (c2 1.5c1 1.5c2x)eⴚ1.5x 10xeⴚ1.5x 7.5x 2eⴚ1.5x.
From this and the second initial condition we have y r (0) c2 1.5c1 0. Hence c2 1.5c1 1.5. This gives
the answer (Fig. 51)
y (1 1.5x)eⴚ1.5x 5x 2eⴚ1.5x (1 1.5x 5x 2)eⴚ1.5x.
The curve begins with a horizontal tangent, crosses the x-axis at x 0.6217 (where 1 1.5x 5x 2 0) and
approaches the axis from below as x increases.
䊏
y
1.0
0.5
0
1
2
3
4
5
x
–0.5
–1.0
Fig. 51. Solution in Example 2
EXAMPLE 3
Application of the Sum Rule (c)
Solve the initial value problem
(7)
y s 2y r 0.75y 2 cos x 0.25 sin x 0.09x,
Solution.
y(0) 2.78,
y r (0) 0.43.
Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous
ODE is
l2 2l 0.75 (l 12 ) (l 32 ) 0
which gives the general solution yh c1eⴚx>2 c2eⴚ3x>2.
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CHAP. 2 Second-Order Linear ODEs
Step 2. Particular solution of the nonhomogeneous ODE. We write yp yp1 yp2 and, following Table 2.1,
(C) and (B),
yp1 K cos x M sin x
and
yp2 K 1x K 0.
Differentiation gives yp1
r K sin x M cos x, yp1
s K cos x M sin x and yp2
r 1, ysp2 0. Substitution
of yp1 into the ODE in (7) gives, by comparing the cosine and sine terms,
K 2M 0.75K 2,
M 2K 0.75M 0.25,
hence K 0 and M 1. Substituting yp2 into the ODE in (7) and comparing the x- and x 0-terms gives
0.75K 1 0.09, 2K 1 0.75K 0 0,
thus
K 1 0.12, K 0 0.32.
Hence a general solution of the ODE in (7) is
y c1eⴚx>2 c2eⴚ3x>2 sin x 0.12x 0.32.
Step 3. Solution of the initial value problem. From y, y r and the initial conditions we obtain
y r (0) 12 c1 32 c2 1 0.12 0.4.
y(0) c1 c2 0.32 2.78,
Hence c1 3.1, c2 0. This gives the solution of the IVP (Fig. 52)
䊏
y 3.1eⴚx>2 sin x 0.12x 0.32.
y
3
2.5
2
1.5
1
0.5
0
2
4
6
8
10 12 14 16 18 20
x
–0.5
Fig. 52.
Solution in Example 3
Stability. The following is important. If (and only if) all the roots of the characteristic
equation of the homogeneous ODE y s ay r by 0 in (4) are negative, or have a negative
real part, then a general solution yh of this ODE goes to 0 as x : , so that the “transient
solution” y yh yp of (4) approaches the “steady-state solution” yp. In this case the
nonhomogeneous ODE and the physical or other system modeled by the ODE are called
stable; otherwise they are called unstable. For instance, the ODE in Example 1 is unstable.
Applications follow in the next two sections.
PROBLEM SET 2.7
1–10
NONHOMOGENEOUS LINEAR ODEs:
GENERAL SOLUTION
Find a (real) general solution. State which rule you are
using. Show each step of your work.
1. y s 5y r 4y 10eⴚ3x
2. 10y s 50y r 57.6y cos x
3. y s 3y r 2y 12x 2
4. y s 9y 18 cos px
5. y s 4y r 4y eⴚx cos x
6. y s y r (p2 14)y eⴚx>2 sin p x
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SEC. 2.8 Modeling: Forced Oscillations. Resonance
7. (D 2 2D 34 I )y 3ex 92 x
8. (3D 2 27I )y 3 cos x cos 3x
9. (D 2 16I )y 9.6e4x 30ex
10. (D 2 2D I )y 2x sin x
11–18
NONHOMOGENEOUS LINEAR
ODEs: IVPs
Solve the initial value problem. State which rule you are
using. Show each step of your calculation in detail.
11. y s 3y 18x 2, y(0) 3, y r (0) 0
12. y s 4y 12 sin 2x, y(0) 1.8, y r (0) 5.0
13. 8y s 6y r y 6 cosh x, y(0) 0.2,
y r (0) 0.05
14. y s 4y r 4y eⴚ2x sin 2x, y(0) 1,
y r (0) 1.5
15. (x 2D 2 3xD 3I )y 3 ln x 4,
y(1) 0, y r (1) 1; yp ln x
16. (D 2 2D)y 6e2x 4eⴚ2x, y(0) 1, y r (0) 6
17. (D 2 0.2D 0.26I)y 1.22e0.5x, y(0) 3.5,
y r (0) 0.35
85
18. (D 2 2D 10I)y 17 sin x 37 sin 3x,
y(0) 6.6, y r (0) 2.2
19. CAS PROJECT. Structure of Solutions of Initial
Value Problems. Using the present method, find,
graph, and discuss the solutions y of initial value
problems of your own choice. Explore effects on
solutions caused by changes of initial conditions.
Graph yp, y, y yp separately, to see the separate
effects. Find a problem in which (a) the part of y
resulting from yh decreases to zero, (b) increases,
(c) is not present in the answer y. Study a problem with
y(0) 0, y r (0) 0. Consider a problem in which
you need the Modification Rule (a) for a simple root,
(b) for a double root. Make sure that your problems
cover all three Cases I, II, III (see Sec. 2.2).
20. TEAM PROJECT. Extensions of the Method of
Undetermined Coefficients. (a) Extend the method
to products of the function in Table 2.1, (b) Extend
the method to Euler–Cauchy equations. Comment on
the practical significance of such extensions.
2.8 Modeling: Forced Oscillations. Resonance
In Sec. 2.4 we considered vertical motions of a mass–spring system (vibration of a mass
m on an elastic spring, as in Figs. 33 and 53) and modeled it by the homogeneous linear
ODE
(1)
my s cy r ky 0.
Here y(t) as a function of time t is the displacement of the body of mass m from rest.
The mass–spring system of Sec. 2.4 exhibited only free motion. This means no external
forces (outside forces) but only internal forces controlled the motion. The internal forces
are forces within the system. They are the force of inertia my s , the damping force cy r
(if c 0), and the spring force ky, a restoring force.
k
Spring
m
Mass
r(t)
c
Dashpot
Fig. 53.
Mass on a spring
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CHAP. 2 Second-Order Linear ODEs
We now extend our model by including an additional force, that is, the external force
r(t), on the right. Then we have
(2*)
my s cy r ky r(t).
Mechanically this means that at each instant t the resultant of the internal forces is in
equilibrium with r(t). The resulting motion is called a forced motion with forcing function
r(t), which is also known as input or driving force, and the solution y(t) to be obtained
is called the output or the response of the system to the driving force.
Of special interest are periodic external forces, and we shall consider a driving force
of the form
r(t) F0 cos vt
(F0 0, v 0).
Then we have the nonhomogeneous ODE
(2)
my s cy r ky F0 cos vt.
Its solution will reveal facts that are fundamental in engineering mathematics and allow
us to model resonance.
Solving the Nonhomogeneous ODE (2)
From Sec. 2.7 we know that a general solution of (2) is the sum of a general solution yh
of the homogeneous ODE (1) plus any solution yp of (2). To find yp, we use the method
of undetermined coefficients (Sec. 2.7), starting from
(3)
yp(t) a cos vt b sin vt.
By differentiating this function (chain rule!) we obtain
ypr va sin vt vb cos vt,
y sp v2a cos vt v2b sin vt.
Substituting yp, ypr , and y sp into (2) and collecting the cosine and the sine terms, we get
[(k mv2)a vcb] cos vt [vca (k mv2)b] sin vt F0 cos vt.
The cosine terms on both sides must be equal, and the coefficient of the sine term
on the left must be zero since there is no sine term on the right. This gives the two
equations
(4)
(k mv2)a vca
vcb
F0
(k mv2)b 0
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SEC. 2.8 Modeling: Forced Oscillations. Resonance
87
for determining the unknown coefficients a and b. This is a linear system. We can solve
it by elimination. To eliminate b, multiply the first equation by k mv2 and the second
by vc and add the results, obtaining
(k mv2)2a v2c2a F0(k mv2).
Similarly, to eliminate a, multiply (the first equation by vc and the second by k mv2
and add to get
v2c2b (k mv2)2b F0vc.
If the factor (k mv2)2 v2c2 is not zero, we can divide by this factor and solve for a
and b,
a F0
k mv2
,
(k mv2)2 v2c2
b F0
vc
.
(k mv2)2 v2c2
If we set 2k>m v0 ( 0) as in Sec. 2.4, then k mv20 and we obtain
(5)
a F0
m(v20 v2)
m 2(v20 v2)2 v2c2
,
b F0
vc
.
m 2(v20 v2)2 v2c2
We thus obtain the general solution of the nonhomogeneous ODE (2) in the form
y(t) yh(t) yp(t).
(6)
Here yh is a general solution of the homogeneous ODE (1) and yp is given by (3) with
coefficients (5).
We shall now discuss the behavior of the mechanical system, distinguishing between
the two cases c 0 (no damping) and c 0 (damping). These cases will correspond to
two basically different types of output.
Case 1. Undamped Forced Oscillations. Resonance
If the damping of the physical system is so small that its effect can be neglected over the
time interval considered, we can set c 0. Then (5) reduces to a F0>[m(v20 v2)]
and b 0. Hence (3) becomes (use v02 k>m)
(7)
yp(t) F0
F0
cos vt cos vt.
2
2
m(v0 v )
k[1 (v>v0)2]
Here we must assume that v2 v02; physically, the frequency v>(2p) [cycles>sec] of
the driving force is different from the natural frequency v0>(2p) of the system, which is
the frequency of the free undamped motion [see (4) in Sec. 2.4]. From (7) and from (4*)
in Sec. 2.4 we have the general solution of the “undamped system”
(8)
y(t) C cos (v0t d) F0
cos vt.
2
m(v0 v2)
We see that this output is a superposition of two harmonic oscillations of the frequencies
just mentioned.
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CHAP. 2 Second-Order Linear ODEs
Resonance. We discuss (7). We see that the maximum amplitude of yp is (put cos vt 1)
(9)
a0 F0
k
r
where
r
1
.
1 (v>v0)2
a0 depends on v and v0. If v : v0, then r and a0 tend to infinity. This excitation of large
oscillations by matching input and natural frequencies (v v0) is called resonance. r is
called the resonance factor (Fig. 54), and from (9) we see that r>k a0>F0 is the ratio
of the amplitudes of the particular solution yp and of the input F0 cos vt. We shall see
later in this section that resonance is of basic importance in the study of vibrating systems.
In the case of resonance the nonhomogeneous ODE (2) becomes
F0
y s v20 y m cos v0t.
(10)
Then (7) is no longer valid, and, from the Modification Rule in Sec. 2.7, we conclude that
a particular solution of (10) is of the form
yp(t) t(a cos v0t b sin v0t).
ρ
1
ω0
ω
Fig. 54. Resonance factor r(v)
By substituting this into (10) we find a 0 and b F0>(2mv0). Hence (Fig. 55)
yp(t) (11)
F0
t sin v0t.
2mv0
yp
t
Fig. 55.
Particular solution in the case of resonance
We see that, because of the factor t, the amplitude of the vibration becomes larger and
larger. Practically speaking, systems with very little damping may undergo large vibrations
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SEC. 2.8 Modeling: Forced Oscillations. Resonance
89
that can destroy the system. We shall return to this practical aspect of resonance later in
this section.
Beats. Another interesting and highly important type of oscillation is obtained if v is
close to v0. Take, for example, the particular solution [see (8)]
y(t) (12)
F0
m(v20 v2)
(cos vt cos v0t)
(v v0).
Using (12) in App. 3.1, we may write this as
y(t) 2F0
m(v20 v2)
sin a
v0 v
2
tb sin a
v0 v
2
tb .
Since v is close to v0, the difference v0 v is small. Hence the period of the last sine
function is large, and we obtain an oscillation of the type shown in Fig. 56, the dashed
curve resulting from the first sine factor. This is what musicians are listening to when
they tune their instruments.
y
t
Fig. 56.
Forced undamped oscillation when the difference of the input
and natural frequencies is small (“beats”)
Case 2. Damped Forced Oscillations
If the damping of the mass–spring system is not negligibly small, we have c 0 and
a damping term cy r in (1) and (2). Then the general solution yh of the homogeneous
ODE (1) approaches zero as t goes to infinity, as we know from Sec. 2.4. Practically,
it is zero after a sufficiently long time. Hence the “transient solution” (6) of (2),
given by y yh yp, approaches the “steady-state solution” yp. This proves the
following.
THEOREM 1
Steady-State Solution
After a sufficiently long time the output of a damped vibrating system under a purely
sinusoidal driving force [see (2)] will practically be a harmonic oscillation whose
frequency is that of the input.
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CHAP. 2 Second-Order Linear ODEs
Amplitude of the Steady-State Solution. Practical Resonance
Whereas in the undamped case the amplitude of yp approaches infinity as v approaches
v0, this will not happen in the damped case. In this case the amplitude will always be
finite. But it may have a maximum for some v depending on the damping constant c.
This may be called practical resonance. It is of great importance because if c is not too
large, then some input may excite oscillations large enough to damage or even destroy
the system. Such cases happened, in particular in earlier times when less was known about
resonance. Machines, cars, ships, airplanes, bridges, and high-rising buildings are vibrating
mechanical systems, and it is sometimes rather difficult to find constructions that are
completely free of undesired resonance effects, caused, for instance, by an engine or by
strong winds.
To study the amplitude of yp as a function of v, we write (3) in the form
yp(t) C* cos (vt h).
(13)
C* is called the amplitude of yp and h the phase angle or phase lag because it measures
the lag of the output behind the input. According to (5), these quantities are
C*(v) 2a 2 b 2 F0
2m
(14)
tan h (v) (v20 v2)2 v2c2
2
,
b
vc
.
2
a
m(v0 v2)
Let us see whether C*(v) has a maximum and, if so, find its location and then its size.
We denote the radicand in the second root in C* by R. Equating the derivative of C* to
zero, we obtain
dC*
1
F0 a R3>2 b [2m 2(v20 v2)(2v) 2vc2].
dv
2
The expression in the brackets [. . .] is zero if
(15)
c2 2m 2(v20 v2)
(v20 k>m).
By reshuffling terms we have
2m 2v2 2m 2v02 c2 2mk c2.
The right side of this equation becomes negative if c2 2mk, so that then (15) has no
real solution and C* decreases monotone as v increases, as the lowest curve in Fig. 57
shows. If c is smaller, c2 2mk, then (15) has a real solution v vmax, where
(15*)
v2max v20 c2
.
2m 2
From (15*) we see that this solution increases as c decreases and approaches v0 as c
approaches zero. See also Fig. 57.
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SEC. 2.8 Modeling: Forced Oscillations. Resonance
91
The size of C*(vmax) is obtained from (14), with v2 v2max given by (15*). For this
v we obtain in the second radicand in (14) from (15*)
2
m 2(v20 v2max)2 c4
4m 2
and
v2max c2 av20 c2
b c2.
2m 2
The sum of the right sides of these two formulas is
(c4 4m 2v20c2 2c4)>(4m 2) c2(4m 2v20 c2)>(4m 2).
Substitution into (14) gives
C*(vmax) (16)
2mF0
c24m 2v20 c2
.
We see that C*(vmax) is always finite when c 0. Furthermore, since the expression
c24m 2v20 c4 c2(4mk c2)
in the denominator of (16) decreases monotone to zero as c2 (2mk) goes to zero, the maximum
amplitude (16) increases monotone to infinity, in agreement with our result in Case 1. Figure 57
shows the amplification C*>F0 (ratio of the amplitudes of output and input) as a function of
v for m 1, k 1, hence v0 1, and various values of the damping constant c.
Figure 58 shows the phase angle (the lag of the output behind the input), which is less
than p>2 when v v0, and greater than p>2 for v v0.
C*
F0
η
π
4
c=0
c = 1/2
c=1
c=2
c = 14_
3
π
__
2
c = 12_
2
c=
1
c=
2
0
1
0
1
2
0
0
ω
Fig. 57. Amplification C*>F0 as a function of
v for m 1, k 1, and various values of the
damping constant c
1
2
ω
Fig. 58. Phase lag h as a function of v for
m 1, k 1, thus v0 1, and various values
of the damping constant c
PROBLEM SET 2.8
STEADY-STATE SOLUTIONS
1. WRITING REPORT. Free and Forced Vibrations.
Write a condensed report of 2–3 pages on the most
important similarities and differences of free and forced
vibrations, with examples of your own. No proofs.
Find the steady-state motion of the mass–spring system
modeled by the ODE. Show the details of your work.
2. Which of Probs. 1–18 in Sec. 2.7 (with x time t)
can be models of mass–spring systems with a harmonic
oscillation as steady-state solution?
4. y s 2.5y r 10y 13.6 sin 4t
3–7
3. y s 6y r 8y 42.5 cos 2t
5. (D 2 D 4.25I )y 22.1 cos 4.5t
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CHAP. 2 Second-Order Linear ODEs
6. (D 2 4D 3I )y cos t 13 cos 3t
7. (4D 2 12D 9I )y 225 75 sin 3t
TRANSIENT SOLUTIONS
8–15
Find the transient motion of the mass–spring system
modeled by the ODE. Show the details of your work.
8. 2y s 4y r 6.5y 4 sin 1.5t
24. Gun barrel. Solve y s y 1 t 2> p2 if 0
t p and 0 if t : ; here, y(0) 0, y r (0) 0. This
models an undamped system on which a force F acts
during some interval of time (see Fig. 59), for instance,
the force on a gun barrel when a shell is fired, the barrel
being braked by heavy springs (and then damped by a
dashpot, which we disregard for simplicity). Hint: At p
both y and y r must be continuous.
9. y s 3y r 3.25y 3 cos t 1.5 sin t
m=1
F
10. y s 16y 56 cos 4t
k=1
1
F = 1 – t2/π2
11. (D 2 2I )y cos 12t sin 12t
π
12. (D 2D 5I )y 4 cos t 8 sin t
F=0
t
2
13. (D 2 I )y cos vt, v2 1
14. (D 2 I )y 5eⴚt cos t
15. (D 2 4D 8I )y 2 cos 2t sin 2t
INITIAL VALUE PROBLEMS
16–20
Find the motion of the mass–spring system modeled by the
ODE and the initial conditions. Sketch or graph the solution
curve. In addition, sketch or graph the curve of y yp to
see when the system practically reaches the steady state.
16. y s 25y 24 sin t, y(0) 1,
17.
y r (0) 1
Fig. 59.
Problem 24
25. CAS EXPERIMENT. Undamped Vibrations.
(a) Solve the initial value problem y s y cos vt,
v2 1, y(0) 0, y r (0) 0. Show that the solution
can be written
y (t) 2
sin [12 (1 v)t] sin [12 (1 v)t].
1 v2
(b) Experiment with the solution by changing v to
see the change of the curves from those for small
v ( 0) to beats, to resonance, and to large values of
v (see Fig. 60).
(D 4I)y sin t 13 sin 3t 15 sin 5t,
3
y(0) 0, y (0) 35
2
2
r
1
18. (D 8D 17I )y 474.5 sin 0.5t, y(0) 5.4,
y r (0) 9.4
19. (D 2 2D 2I )y eⴚt>2 sin 12 t,
y r (0) 1
10π
y(0) 0,
20. (D 2 5I )y cos pt sin pt, y(0) 0, y r (0) 0
ω = 0.2
21. Beats. Derive the formula after (12) from (12). Can
we have beats in a damped system?
10
22. Beats. Solve y s 25y 99 cos 4.9t, y(0) 2,
y r (0) 0. How does the graph of the solution change
if you change (a) y(0), (b) the frequency of the driving
force?
23. TEAM EXPERIMENT. Practical Resonance.
(a) Derive, in detail, the crucial formula (16).
(b) By considering dC*>dc show that C*(vmax) increases as c ( 12mk) decreases.
(c) Illustrate practical resonance with an ODE of your
own in which you vary c, and sketch or graph
corresponding curves as in Fig. 57.
(d) Take your ODE with c fixed and an input of two
terms, one with frequency close to the practical
resonance frequency and the other not. Discuss and
sketch or graph the output.
(e) Give other applications (not in the book) in which
resonance is important.
20π
–1
20π
–10
ω = 0.9
0.04
10π
–0.04
ω=6
Fig. 60.
Typical solution curves in CAS Experiment 25
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SEC. 2.9 Modeling: Electric Circuits
93
2.9 Modeling: Electric Circuits
Designing good models is a task the computer cannot do. Hence setting up models has
become an important task in modern applied mathematics. The best way to gain experience
in successful modeling is to carefully examine the modeling process in various fields and
applications. Accordingly, modeling electric circuits will be profitable for all students,
not just for electrical engineers and computer scientists.
Figure 61 shows an RLC-circuit, as it occurs as a basic building block of large electric
networks in computers and elsewhere. An RLC-circuit is obtained from an RL-circuit by
adding a capacitor. Recall Example 2 on the RL-circuit in Sec. 1.5: The model of the
RL-circuit is LI r RI E(t). It was obtained by KVL (Kirchhoff’s Voltage Law)7 by
equating the voltage drops across the resistor and the inductor to the EMF (electromotive
force). Hence we obtain the model of the RLC-circuit simply by adding the voltage drop
Q> C across the capacitor. Here, C F (farads) is the capacitance of the capacitor. Q coulombs
is the charge on the capacitor, related to the current by
I(t) dQ
,
dt
equivalently
冮
Q(t) I(t) dt.
See also Fig. 62. Assuming a sinusoidal EMF as in Fig. 61, we thus have the model of
the RLC-circuit
C
R
L
E(t) = E0 sin ω
ωt
Fig. 61. RLC-circuit
Name
Symbol
Notation
Unit
Voltage Drop
RI
dI
L
dt
Q/C
Ohm’s Resistor
R
Ohm’s Resistance
ohms ( )
Inductor
L
Inductance
henrys (H)
Capacitor
C
Capacitance
farads (F)
Fig. 62.
Elements in an RLC-circuit
7
GUSTAV ROBERT KIRCHHOFF (1824–1887), German physicist. Later we shall also need Kirchhoff’s
Current Law (KCL):
At any point of a circuit, the sum of the inflowing currents is equal to the sum of the outflowing currents.
The units of measurement of electrical quantities are named after ANDRÉ MARIE AMPÈRE (1775–1836),
French physicist, CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer,
MICHAEL FARADAY (1791–1867), English physicist, JOSEPH HENRY (1797–1878), American physicist,
GEORG SIMON OHM (1789–1854), German physicist, and ALESSANDRO VOLTA (1745–1827), Italian
physicist.
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CHAP. 2 Second-Order Linear ODEs
(1 r )
LI r RI 冮
1
I dt E(t) E 0 sin vt.
C
This is an “integro-differential equation.” To get rid of the integral, we differentiate (1 r )
with respect to t, obtaining
(1)
LI s RI r 1
I E r (t) E 0v cos vt.
C
This shows that the current in an RLC-circuit is obtained as the solution of this
nonhomogeneous second-order ODE (1) with constant coefficients.
In connection with initial value problems, we shall occasionally use
(1 s )
LQ s RQ s 1
Q E(t),
C
obtained from (1 r ) and I Q r .
Solving the ODE (1) for the Current in an RLC-Circuit
A general solution of (1) is the sum I Ih Ip, where Ih is a general solution of the
homogeneous ODE corresponding to (1) and Ip is a particular solution of (1). We first
determine Ip by the method of undetermined coefficients, proceeding as in the previous
section. We substitute
(2)
Ip a cos vt b sin vt
Ipr v(a sin vt b cos vt)
Ips v2(a cos vt b sin vt)
into (1). Then we collect the cosine terms and equate them to E 0v cos vt on the right,
and we equate the sine terms to zero because there is no sine term on the right,
Lv2(a) Rvb a>C E 0v
(Cosine terms)
Lv2(b) Rv(a) b>C 0
(Sine terms).
Before solving this system for a and b, we first introduce a combination of L and C, called
the reactance
(3)
S vL 1
.
vC
Dividing the previous two equations by v, ordering them, and substituting S gives
Sa Rb E 0
Ra Sb 0.
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SEC. 2.9 Modeling: Electric Circuits
95
We now eliminate b by multiplying the first equation by S and the second by R, and
adding. Then we eliminate a by multiplying the first equation by R and the second by
S, and adding. This gives
(S 2 R2)a E 0 S,
(R2 S 2)b E 0 R.
We can solve for a and b,
a
(4)
E 0 S
R S
2
2
,
b
E0 R
R S2
2
.
Equation (2) with coefficients a and b given by (4) is the desired particular solution Ip of
the nonhomogeneous ODE (1) governing the current I in an RLC-circuit with sinusoidal
electromotive force.
Using (4), we can write Ip in terms of “physically visible” quantities, namely, amplitude
I0 and phase lag u of the current behind the EMF, that is,
Ip(t) I0 sin (vt u)
(5)
where [see (14) in App. A3.1]
I0 2a 2 b 2 E0
2R S
2
2
,
tan u a
b
S
.
R
The quantity 2R2 S 2 is called the impedance. Our formula shows that the impedance
equals the ratio E 0>I0. This is somewhat analogous to E>I R (Ohm’s law) and, because
of this analogy, the impedance is also known as the apparent resistance.
A general solution of the homogeneous equation corresponding to (1) is
Ih c1el1t c2el2t
where l1 and l2 are the roots of the characteristic equation
l2 R
1
l
0.
L
LC
We can write these roots in the form l1 a b and l2 a b, where
a
R
,
2L
b
R2
1
1
4L
R2 .
2
LC
2L B
C
B 4L
Now in an actual circuit, R is never zero (hence R 0). From this it follows that Ih
approaches zero, theoretically as t : , but practically after a relatively short time. Hence
the transient current I Ih Ip tends to the steady-state current Ip, and after some time
the output will practically be a harmonic oscillation, which is given by (5) and whose
frequency is that of the input (of the electromotive force).
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CHAP. 2 Second-Order Linear ODEs
EXAMPLE 1
RLC-Circuit
Find the current I(t) in an RLC-circuit with R 11 (ohms), L 0.1 H (henry), C 10ⴚ2 F (farad), which
is connected to a source of EMF E(t) 110 sin (60 # 2pt) 110 sin 377 t (hence 60 Hz 60 cycles> sec, the
usual in the U.S. and Canada; in Europe it would be 220 V and 50 Hz). Assume that current and capacitor
charge are 0 when t 0.
Solution. Step 1. General solution of the homogeneous ODE. Substituting R, L, C and the derivative E r (t)
into (1), we obtain
0.1I s 11I r 100I 110 # 377 cos 377t.
Hence the homogeneous ODE is 0.1I s 11I r 100I 0. Its characteristic equation is
0.1l2 11l 100 0.
The roots are l1 10 and l2 100. The corresponding general solution of the homogeneous ODE is
Ih(t) c1eⴚ10t c2eⴚ100t.
Step 2. Particular solution Ip of (1). We calculate the reactance S 37.7 0.3 37.4 and the steady-state
current
Ip(t) a cos 377t b sin 377t
with coefficients obtained from (4) (and rounded)
a
110 # 37.4
11 37.4
2
2
2.71,
b
110 # 11
112 37.42
0.796.
Hence in our present case, a general solution of the nonhomogeneous ODE (1) is
(6)
I(t) c1eⴚ10t c2eⴚ100t 2.71 cos 377t 0.796 sin 377t.
Step 3. Particular solution satisfying the initial conditions. How to use Q(0) 0? We finally determine c1
and c2 from the in initial conditions I(0) 0 and Q(0) 0. From the first condition and (6) we have
(7)
I(0) c1 c2 2.71 0,
c2 2.71 c1.
hence
We turn to Q(0) 0. The integral in (1 r ) equals 兰 I dt Q(t); see near the beginning of this section. Hence for
t 0, Eq. (1 r ) becomes
LI r (0) R # 0 0,
so that
I r (0) 0.
Differentiating (6) and setting t 0, we thus obtain
I r (0) 10c1 100c2 0 0.796 # 377 0,
hence by (7),
10c1 100(2.71 c1) 300.1.
The solution of this and (7) is c1 0.323, c2 3.033. Hence the answer is
I(t) 0.323eⴚ10t 3.033eⴚ100t 2.71 cos 377t 0.796 sin 377t .
You may get slightly different values depending on the rounding. Figure 63 shows I(t) as well as Ip(t), which
practically coincide, except for a very short time near t 0 because the exponential terms go to zero very rapidly.
Thus after a very short time the current will practically execute harmonic oscillations of the input frequency
60 Hz 60 cycles> sec. Its maximum amplitude and phase lag can be seen from (5), which here takes the form
Ip(t) 2.824 sin (377t 1.29).
䊏
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SEC. 2.9 Modeling: Electric Circuits
97
y
I(t)
3
2
1
0
0.01
0.02
0.03
0.04
0.05
t
–1
–2
–3
Fig. 63.
Transient (upper curve) and steady-state currents in Example 1
Analogy of Electrical and Mechanical Quantities
Entirely different physical or other systems may have the same mathematical model.
For instance, we have seen this from the various applications of the ODE y r ky in
Chap. 1. Another impressive demonstration of this unifying power of mathematics is
given by the ODE (1) for an electric RLC-circuit and the ODE (2) in the last section for
a mass–spring system. Both equations
LI s RI r 1
I E 0v cos vt
C
and
my s cy r ky F0 cos vt
are of the same form. Table 2.2 shows the analogy between the various quantities involved.
The inductance L corresponds to the mass m and, indeed, an inductor opposes a change
in current, having an “inertia effect” similar to that of a mass. The resistance R corresponds
to the damping constant c, and a resistor causes loss of energy, just as a damping dashpot
does. And so on.
This analogy is strictly quantitative in the sense that to a given mechanical system we
can construct an electric circuit whose current will give the exact values of the displacement
in the mechanical system when suitable scale factors are introduced.
The practical importance of this analogy is almost obvious. The analogy may be used
for constructing an “electrical model” of a given mechanical model, resulting in substantial
savings of time and money because electric circuits are easy to assemble, and electric
quantities can be measured much more quickly and accurately than mechanical ones.
Table 2.2 Analogy of Electrical and Mechanical Quantities
Electrical System
Inductance L
Resistance R
Reciprocal 1> C of capacitance
Derivative E 0v cos vt of
}
electromotive force
Current I(t)
Mechanical System
Mass m
Damping constant c
Spring modulus k
Driving force F0 cos vt
Displacement y(t)
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CHAP. 2 Second-Order Linear ODEs
Related to this analogy are transducers, devices that convert changes in a mechanical
quantity (for instance, in a displacement) into changes in an electrical quantity that can
be monitored; see Ref. [GenRef11] in App. 1.
PROBLEM SET 2.9
1–6
RLC-CIRCUITS: SPECIAL CASES
1. RC-Circuit. Model the RC-circuit in Fig. 64. Find the
current due to a constant E.
4. RL-Circuit. Solve Prob. 3 when E E 0 sin vt and R,
L, E 0, and are arbitrary. Sketch a typical solution.
Current I(t)
2
R
1.5
1
E(t)
0.5
4π
C
Fig. 64. RC-circuit
Fig. 68.
c
Typical current I eⴚ0.1t sin (t 41 p)
in Problem 4
5. LC-Circuit. This is an RLC-circuit with negligibly
small R (analog of an undamped mass–spring system).
Find the current when L 0.5 H, C 0.005 F, and
E sin t V, assuming zero initial current and charge.
t
Current 1 in Problem 1
2. RC-Circuit. Solve Prob. 1 when E E 0 sin vt and
R, C, E 0, and v are arbitrary.
3. RL-Circuit. Model the RL-circuit in Fig. 66. Find a
general solution when R, L, E are any constants. Graph
or sketch solutions when L 0.25 H, R 10 , and
E 48 V.
C
L
E(t)
Fig. 69.
R
LC-circuit
6. LC-Circuit. Find the current when L 0.5 H,
C 0.005 F, E 2t 2 V, and initial current and charge
zero.
E(t)
7–18
L
Fig. 66.
t
–1
Current I(t)
Fig. 65.
12π
8π
–0.5
GENERAL RLC-CIRCUITS
7. Tuning. In tuning a stereo system to a radio station,
we adjust the tuning control (turn a knob) that changes
C (or perhaps L) in an RLC-circuit so that the amplitude
of the steady-state current (5) becomes maximum. For
what C will this happen?
RL-circuit
Current I(t)
5
4
3
8–14
Find the steady-state current in the RLC-circuit
in Fig. 61 for the given data. Show the details of your work.
2
1
0
0.02
0.04
0.06
0.08
0.1
Fig. 67. Currents in Problem 3
t
8. R 4 , L 0.5 H, C 0.1 F, E 500 sin 2t V
9. R 4 , L 0.1 H, C 0.05 F, E 110 V
1
10. R 2 , L 1 H, C 20
F, E 157 sin 3t V
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SEC. 2.10 Solution by Variation of Parameters
99
1
11. R 12 , L 0.4 H, C 80
F,
E 220 sin 10t V
12. R 0.2
, L 0.1 H, C 2 F, E 220 sin 314t V
# 10ⴚ3 F,
13. R 12, L 1.2 H, C 20
3
E 12,000 sin 25t V
14. Prove the claim in the text that if R 0 (hence R 0),
then the transient current approaches Ip as t : .
15. Cases of damping. What are the conditions for an
RLC-circuit to be (I) overdamped, (II) critically damped,
(III) underdamped? What is the critical resistance Rcrit
(the analog of the critical damping constant 2 1mk)?
16–18
Solve the initial value problem for the RLCcircuit in Fig. 61 with the given data, assuming zero initial
current and charge. Graph or sketch the solution. Show the
details of your work.
16. R 8 , L 0.2 H, C 12.5 # 10ⴚ3 F,
E 100 sin 10t V
17. R 6 , L 1 H, C 0.04 F,
E 600 (cos t 4 sin t) V
18. R 18 , L 1 H, C 12.5 # 10ⴚ3 F,
E 820 cos 10t V
19. WRITING REPORT. Mechanic-Electric Analogy.
Explain Table 2.2 in a 1–2 page report with examples,
e.g., the analog (with L 1 H) of a mass–spring system
of mass 5 kg, damping constant 10 kg>sec, spring constant
60 kg>sec2, and driving force 220 cos 10t kg>sec.
~
~
20. Complex Solution Method. Solve LI s RI r ~
ivt
I >C E 0e , i 11, by substituting Ip Keivt
(K unknown) and its derivatives and taking the real
~
part Ip of the solution I p . Show agreement with (2), (4).
ivt
Hint: Use (11) e cos vt i sin vt; cf. Sec. 2.2,
and i 2 1.
2.10 Solution by Variation of Parameters
We continue our discussion of nonhomogeneous linear ODEs, that is
(1)
y s p(x)y r q(x)y r (x).
In Sec. 2.6 we have seen that a general solution of (1) is the sum of a general solution yh
of the corresponding homogeneous ODE and any particular solution yp of (1). To obtain yp
when r (x) is not too complicated, we can often use the method of undetermined coefficients,
as we have shown in Sec. 2.7 and applied to basic engineering models in Secs. 2.8 and 2.9.
However, since this method is restricted to functions r (x) whose derivatives are of a form
similar to r (x) itself (powers, exponential functions, etc.), it is desirable to have a method valid
for more general ODEs (1), which we shall now develop. It is called the method of variation
of parameters and is credited to Lagrange (Sec. 2.1). Here p, q, r in (1) may be variable
(given functions of x), but we assume that they are continuous on some open interval I.
Lagrange’s method gives a particular solution yp of (1) on I in the form
(2)
yp(x) y1
冮 W dx y 冮 W dx
y2r
y1r
2
where y1, y2 form a basis of solutions of the corresponding homogeneous ODE
(3)
y s p(x)y r q(x)y 0
on I, and W is the Wronskian of y1, y2,
(4)
W y1y2r y2y1r
(see Sec. 2.6).
CAUTION! The solution formula (2) is obtained under the assumption that the ODE
is written in standard form, with y s as the first term as shown in (1). If it starts with
f (x)y s , divide first by f (x).
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CHAP. 2 Second-Order Linear ODEs
The integration in (2) may often cause difficulties, and so may the determination of
y1, y 2 if (1) has variable coefficients. If you have a choice, use the previous method. It is
simpler. Before deriving (2) let us work an example for which you do need the new
method. (Try otherwise.)
EXAMPLE 1
Method of Variation of Parameters
Solve the nonhomogeneous ODE
y s y sec x 1
.
cos x
A basis of solutions of the homogeneous ODE on any interval is y1 cos x, y2 sin x. This gives
the Wronskian
Solution.
W( y1, y2) cos x cos x sin x (sin x) 1.
From (2), choosing zero constants of integration, we get the particular solution of the given ODE
冮
冮
yp cos x sin x sec x dx sin x cos x sec x dx
cos x ln ƒ cos x ƒ x sin x
(Fig. 70)
Figure 70 shows yp and its first term, which is small, so that x sin x essentially determines the shape of the curve
of yp. (Recall from Sec. 2.8 that we have seen x sin x in connection with resonance, except for notation.) From
yp and the general solution yh c1y1 c2y2 of the homogeneous ODE we obtain the answer
y yh yp (c1 ln ƒ cos x ƒ ) cos x (c2 x) sin x.
Had we included integration constants c1, c2 in (2), then (2) would have given the additional
c1 cos x c2 sin x c1y1 c2y2, that is, a general solution of the given ODE directly from (2). This will
䊏
always be the case.
y
10
5
0
2
4
6
8 10 12 x
–5
–10
Fig. 70. Particular solution yp and its first term in Example 1
Idea of the Method. Derivation of (2)
What idea did Lagrange have? What gave the method the name? Where do we use the
continuity assumptions?
The idea is to start from a general solution
yh(x) c1y1(x) c2y2(x)
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SEC. 2.10 Solution by Variation of Parameters
101
of the homogeneous ODE (3) on an open interval I and to replace the constants (“the
parameters”) c1 and c2 by functions u(x) and v(x); this suggests the name of the method.
We shall determine u and v so that the resulting function
yp(x) u(x)y1(x) v(x)y2(x)
(5)
is a particular solution of the nonhomogeneous ODE (1). Note that yh exists by Theorem
3 in Sec. 2.6 because of the continuity of p and q on I. (The continuity of r will be used
later.)
We determine u and v by substituting (5) and its derivatives into (1). Differentiating (5),
we obtain
ypr u r y1 uy1r v r y2 vy2r .
Now yp must satisfy (1). This is one condition for two functions u and v. It seems plausible
that we may impose a second condition. Indeed, our calculation will show that we can
determine u and v such that yp satisfies (1) and u and v satisfy as a second condition the
equation
u r y1 v r y2 0.
(6)
This reduces the first derivative ypr to the simpler form
ypr uy1r vy2r .
(7)
Differentiating (7), we obtain
yps u r y1r uy1s v r y2r vy2s .
(8)
We now substitute yp and its derivatives according to (5), (7), (8) into (1). Collecting
terms in u and terms in v, we obtain
u( y1s py1r qy1) v( y2s py2r qy2) u r y1r v r y2r r.
Since y1 and y2 are solutions of the homogeneous ODE (3), this reduces to
(9a)
u r y1r v r y2r r.
Equation (6) is
(9b)
u r y1 v r y2 0.
This is a linear system of two algebraic equations for the unknown functions u r and v r .
We can solve it by elimination as follows (or by Cramer’s rule in Sec. 7.6). To eliminate
v r , we multiply (9a) by y2 and (9b) by y2r and add, obtaining
u r (y1y2r y2y1r ) y2r,
thus
u r W y2r.
Here, W is the Wronskian (4) of y1, y2. To eliminate u r we multiply (9a) by y1, and (9b)
by y1r and add, obtaining
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CHAP. 2 Second-Order Linear ODEs
v r (y1y 2r y2y1r ) y1r,
v r W y1r.
thus
Since y1, y 2 form a basis, we have W 0 (by Theorem 2 in Sec. 2.6) and can divide by W,
(10)
y2r
,
W
vr 冮 W dx,
v
ur y1r
.
W
By integration,
u
y2r
冮 W dx.
y1r
These integrals exist because r (x) is continuous. Inserting them into (5) gives (2) and
completes the derivation.
䊏
PROBLEM SET 2.10
1–13
GENERAL SOLUTION
Solve the given nonhomogeneous linear ODE by variation
of parameters or undetermined coefficients. Show the
details of your work.
1. y s 9y sec 3x
2. y s 9y csc 3x
3. x 2y s 2xy r 2y x 3 sin x
4. y s 4y r 5y e2x csc x
5. y s y cos x sin x
6. (D 2 6D 9I )y 16eⴚ3x>(x 2 1)
7. (D 2 4D 4I )y 6e2x>x 4
8. (D 2 4I )y cosh 2x
9. (D 2 2D I )y 35x 3>2ex
10. (D 2 2D 2I )y 4eⴚx sec3 x
11. (x 2D 2 4xD 6I )y 21x ⴚ4
12. (D 2 I )y 1>cosh x
13. (x 2D 2 xD 9I )y 48x 5
14. TEAM PROJECT. Comparison of Methods. Invention. The undetermined-coefficient method should be
used whenever possible because it is simpler. Compare
it with the present method as follows.
(a) Solve y s 4y r 3y 65 cos 2x by both methods,
showing all details, and compare.
(b) Solve y s 2y r y r1 r2, r1 35x 3>2ex r2 x 2 by applying each method to a suitable function on
the right.
(c) Experiment to invent an undetermined-coefficient
method for nonhomogeneous Euler–Cauchy equations.
CHAPTER 2 REVIEW QUESTIONS AND PROBLEMS
1. Why are linear ODEs preferable to nonlinear ones in
modeling?
2. What does an initial value problem of a second-order
ODE look like? Why must you have a general solution
to solve it?
3. By what methods can you get a general solution of a
nonhomogeneous ODE from a general solution of a
homogeneous one?
4. Describe applications of ODEs in mechanical systems.
What are the electrical analogs of the latter?
5. What is resonance? How can you remove undesirable
resonance of a construction, such as a bridge, a ship,
or a machine?
6. What do you know about existence and uniqueness of
solutions of linear second-order ODEs?
7–18
GENERAL SOLUTION
Find a general solution. Show the details of your calculation.
7. 4y s 32y r 63y 0
8. y s y r 12y 0
9. y s 6y r 34y 0
10. y s 0.20y r 0.17y 0
11. (100D 2 160D 64I )y 0
12. (D 2 4pD 4p2I )y 0
13. (x 2D 2 2xD 12I )y 0
14. (x 2D 2 xD 9I )y 0
15. (2D 2 3D 2I )y 13 2x 2
16. (D 2 2D 2I )y 3eⴚx cos 2x
17. (4D 2 12D 9I )y 2e1.5x
18. yy s 2y r 2
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Summary of Chapter 2
19–22
103
INITIAL VALUE PROBLEMS
Solve the problem, showing the details of your work.
Sketch or graph the solution.
19. y s 16y 17ex, y(0) 6, y r (0) 2
20. y s 3y r 2y 10 sin x, y(0) 1, y r (0) 6
21. (x 2D 2 xD I )y 16x 3, y(1) 1, y r (1) 1
22. (x 2D 2 15xD 49I )y 0, y(1) 2,
y r (1) 11
23–30
27. Find an electrical analog of the mass–spring system
with mass 4 kg, spring constant 10 kg>sec2, damping
constant 20 kg> sec, and driving force 100 sin 4t nt.
28. Find the motion of the mass–spring system in Fig. 72
with mass 0.125 kg, damping 0, spring constant
1.125 kg>sec2, and driving force cos t 4 sin t nt, assuming zero initial displacement and velocity. For what
frequency of the driving force would you get resonance?
APPLICATIONS
23. Find the steady-state current in the RLC-circuit in Fig. 71
when R 2 k (2000 ), L 1 H, C 4 # 10ⴚ3 F, and
E 110 sin 415t V (66 cycles> sec).
24. Find a general solution of the homogeneous linear
ODE corresponding to the ODE in Prob. 23.
25. Find the steady-state current in the RLC-circuit
in Fig. 71 when R 50 , L 30 H, C 0.025 F,
E 200 sin 4t V.
C
Spring
m
Mass
c
Dashpot
Fig. 72. Mass–spring system
29. Show that the system in Fig. 72 with m 4, c 0,
k 36, and driving force 61 cos 3.1t exhibits beats.
Hint: Choose zero initial conditions.
L
R
k
E(t )
Fig. 71. RLC-circuit
26. Find the current in the RLC-circuit in Fig. 71
when R 40 , L 0.4 H, C 10ⴚ4 F, E 220 sin 314t V (50 cycles> sec).
SUMMARY OF CHAPTER
30. In Fig. 72, let m 1 kg, c 4 kg> sec, k 24 kg>sec2,
and r(t) 10 cos vt nt. Determine w such that you
get the steady-state vibration of maximum possible
amplitude. Determine this amplitude. Then find the
general solution with this v and check whether the results
are in agreement.
2
Second-Order Linear ODEs
Second-order linear ODEs are particularly important in applications, for instance,
in mechanics (Secs. 2.4, 2.8) and electrical engineering (Sec. 2.9). A second-order
ODE is called linear if it can be written
(1)
y s p(x)y r q(x)y r (x)
(Sec. 2.1).
(If the first term is, say, f (x)y s , divide by f (x) to get the “standard form” (1) with
y s as the first term.) Equation (1) is called homogeneous if r (x) is zero for all x
considered, usually in some open interval; this is written r (x) ⬅ 0. Then
(2)
y s p(x)y r q(x)y 0.
Equation (1) is called nonhomogeneous if r (x) [ 0 (meaning r (x) is not zero for
some x considered).
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CHAP. 2 Second-Order Linear ODEs
For the homogeneous ODE (2) we have the important superposition principle (Sec.
2.1) that a linear combination y ky1 ly2 of two solutions y1, y2 is again a solution.
Two linearly independent solutions y1, y2 of (2) on an open interval I form a basis
(or fundamental system) of solutions on I. and y c1y1 c2y2 with arbitrary
constants c1, c2 a general solution of (2) on I. From it we obtain a particular
solution if we specify numeric values (numbers) for c1 and c2, usually by prescribing
two initial conditions
y(x 0) K 0,
(3)
y r (x 0) K 1
(x 0, K 0, K 1 given numbers; Sec. 2.1).
(2) and (3) together form an initial value problem. Similarly for (1) and (3).
For a nonhomogeneous ODE (1) a general solution is of the form
y yh yp
(4)
(Sec. 2.7).
Here yh is a general solution of (2) and yp is a particular solution of (1). Such a yp
can be determined by a general method (variation of parameters, Sec. 2.10) or in
many practical cases by the method of undetermined coefficients. The latter applies
when (1) has constant coefficients p and q, and r (x) is a power of x, sine, cosine,
etc. (Sec. 2.7). Then we write (1) as
y s ay r by r (x)
(5)
(Sec. 2.7).
The corresponding homogeneous ODE y r ay r by 0 has solutions y elx,
where l is a root of
l2 al b 0.
(6)
Hence there are three cases (Sec. 2.2):
Case
Type of Roots
General Solution
I
II
III
Distinct real l1, l2
Double 12 a
Complex 12 a iv*
y c1el1x c2el2x
y (c1 c2x)e ax>2
y eⴚax>2(A cos v*x B sin v*x)
Here v* is used since v is needed in driving forces.
Important applications of (5) in mechanical and electrical engineering in connection
with vibrations and resonance are discussed in Secs. 2.4, 2.7, and 2.8.
Another large class of ODEs solvable “algebraically” consists of the Euler–Cauchy
equations
(7)
x 2y s axy r by 0
(Sec. 2.5).
These have solutions of the form y x m, where m is a solution of the auxiliary equation
(8)
m 2 (a 1)m b 0.
Existence and uniqueness of solutions of (1) and (2) is discussed in Secs. 2.6
and 2.7, and reduction of order in Sec. 2.1.
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CHAPTER
3
Higher Order Linear ODEs
The concepts and methods of solving linear ODEs of order n ⫽ 2 extend nicely to linear
ODEs of higher order n, that is, n ⫽ 3, 4, etc. This shows that the theory explained in
Chap. 2 for second-order linear ODEs is attractive, since it can be extended in a
straightforward way to arbitrary n. We do so in this chapter and notice that the formulas
become more involved, the variety of roots of the characteristic equation (in Sec. 3.2)
becomes much larger with increasing n, and the Wronskian plays a more prominent role.
The concepts and methods of solving second-order linear ODEs extend readily to linear
ODEs of higher order.
This chapter follows Chap. 2 naturally, since the results of Chap. 2 can be readily
extended to that of Chap. 3.
Prerequisite: Secs. 2.1, 2.2, 2.6, 2.7, 2.10.
References and Answers to Problems: App. 1 Part A, and App. 2.
3.1 Homogeneous Linear ODEs
Recall from Sec. 1.1 that an ODE is of nth order if the nth derivative y (n) ⫽ d ny>dx n of
the unknown function y(x) is the highest occurring derivative. Thus the ODE is of the form
F (x, y, y r , Á , y (n)) ⫽ 0
where lower order derivatives and y itself may or may not occur. Such an ODE is called
linear if it can be written
(1)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ r (x).
(For n ⫽ 2 this is (1) in Sec. 2.1 with p1 ⫽ p and p0 ⫽ q.) The coefficients p0, Á , pnⴚ1
and the function r on the right are any given functions of x, and y is unknown. y (n) has
coefficient 1. We call this the standard form. (If you have pn(x)y (n), divide by pn(x)
to get this form.) An nth-order ODE that cannot be written in the form (1) is called
nonlinear.
If r (x) is identically zero, r (x) ⬅ 0 (zero for all x considered, usually in some open
interval I), then (1) becomes
(2)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ 0
105
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CHAP. 3 Higher Order Linear ODEs
and is called homogeneous. If r (x) is not identically zero, then the ODE is called
nonhomogeneous. This is as in Sec. 2.1.
A solution of an nth-order (linear or nonlinear) ODE on some open interval I is a
function y ⫽ h(x) that is defined and n times differentiable on I and is such that the ODE
becomes an identity if we replace the unknown function y and its derivatives by h and its
corresponding derivatives.
Sections 3.1–3.2 will be devoted to homogeneous linear ODEs and Section 3.3 to
nonhomogeneous linear ODEs.
Homogeneous Linear ODE: Superposition Principle,
General Solution
The basic superposition or linearity principle of Sec. 2.1 extends to nth order
homogeneous linear ODEs as follows.
THEOREM 1
Fundamental Theorem for the Homogeneous Linear ODE (2)
For a homogeneous linear ODE (2), sums and constant multiples of solutions on
some open interval I are again solutions on I. (This does not hold for a
nonhomogeneous or nonlinear ODE!)
The proof is a simple generalization of that in Sec. 2.1 and we leave it to the student.
Our further discussion parallels and extends that for second-order ODEs in Sec. 2.1.
So we next define a general solution of (2), which will require an extension of linear
independence from 2 to n functions.
DEFINITION
General Solution, Basis, Particular Solution
A general solution of (2) on an open interval I is a solution of (2) on I of the form
(3)
y(x) ⫽ c1 y1(x) ⫹ Á ⫹ cn yn(x)
(c1, Á , cn arbitrary)
where y1, Á , yn is a basis (or fundamental system) of solutions of (2) on I; that
is, these solutions are linearly independent on I, as defined below.
A particular solution of (2) on I is obtained if we assign specific values to the
n constants c1, Á , cn in (3).
DEFINITION
Linear Independence and Dependence
Consider n functions y1(x), Á , yn(x) defined on some interval I.
These functions are called linearly independent on I if the equation
(4)
k1 y1(x) ⫹ Á ⫹ k n yn(x) ⫽ 0
on I
implies that all k1, Á , k n are zero. These functions are called linearly dependent
on I if this equation also holds on I for some k1, Á , k n not all zero.
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SEC. 3.1 Homogeneous Linear ODEs
107
If and only if y1, Á , yn are linearly dependent on I, we can express (at least) one of
these functions on I as a “linear combination” of the other n ⫺ 1 functions, that is, as a
sum of those functions, each multiplied by a constant (zero or not). This motivates the
term “linearly dependent.” For instance, if (4) holds with k 1 ⫽ 0, we can divide by k 1
and express y1 as the linear combination
y1 ⫽ ⫺
1
(k 2 y2 ⫹ Á ⫹ k n yn).
k1
Note that when n ⫽ 2, these concepts reduce to those defined in Sec. 2.1.
EXAMPLE 1
Linear Dependence
Show that the functions y1 ⫽ x 2, y2 ⫽ 5x, y3 ⫽ 2x are linearly dependent on any interval.
Solution.
EXAMPLE 2
䊏
y2 ⫽ 0y1 ⫹ 2.5y3. This proves linear dependence on any interval.
Linear Independence
Show that y1 ⫽ x, y2 ⫽ x 2, y3 ⫽ x 3 are linearly independent on any interval, for instance, on ⫺1 ⬉ x ⬉ 2.
Solution.
Equation (4) is k 1x ⫹ k 2x 2 ⫹ k 3x 3 ⫽ 0. Taking (a) x ⫽ ⫺1, (b) x ⫽ 1, (c) x ⫽ 2, we get
(a) ⫺k 1 ⫹ k 2 ⫺ k 3 ⫽ 0,
(b) k 1 ⫹ k 2 ⫹ k 3 ⫽ 0,
(c) 2k 1 ⫹ 4k 2 ⫹ 8k 3 ⫽ 0.
k 2 ⫽ 0 from (a) ⫹ (b). Then k 3 ⫽ 0 from (c) ⫺2(b). Then k 1 ⫽ 0 from (b). This proves linear independence.
A better method for testing linear independence of solutions of ODEs will soon be explained.
䊏
EXAMPLE 3
General Solution. Basis
Solve the fourth-order ODE
y iv ⫺ 5y s ⫹ 4y ⫽ 0
Solution.
(where y iv ⫽ d 4y>dx 4).
As in Sec. 2.2 we substitute y ⫽ elx. Omitting the common factor elx, we obtain the characteristic
equation
l4 ⫺ 5l2 ⫹ 4 ⫽ 0.
This is a quadratic equation in ␮ ⫽ l2, namely,
␮2 ⫺ 5␮ ⫹ 4 ⫽ (␮ ⫺ 1)(␮ ⫺ 4) ⫽ 0.
The roots are ␮ ⫽ 1 and 4. Hence l ⫽ ⫺2, ⫺1, 1, 2. This gives four solutions. A general solution on any
interval is
y ⫽ c1eⴚ2x ⫹ c2eⴚx ⫹ c3ex ⫹ c4e2x
provided those four solutions are linearly independent. This is true but will be shown later.
Initial Value Problem. Existence and Uniqueness
An initial value problem for the ODE (2) consists of (2) and n initial conditions
(5)
y(x 0) ⫽ K 0,
y r (x 0) ⫽ K 1,
Á,
y (nⴚ1)(x 0) ⫽ K nⴚ1
with given x 0 in the open interval I considered, and given K 0, Á , K nⴚ1.
䊏
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CHAP. 3 Higher Order Linear ODEs
In extension of the existence and uniqueness theorem in Sec. 2.6 we now have the
following.
THEOREM 2
Existence and Uniqueness Theorem for Initial Value Problems
If the coefficients p0(x), Á , pnⴚ1(x) of (2) are continuous on some open interval I
and x 0 is in I, then the initial value problem (2), (5) has a unique solution y(x) on I.
Existence is proved in Ref. [A11] in App. 1. Uniqueness can be proved by a slight
generalization of the uniqueness proof at the beginning of App. 4.
EXAMPLE 4
Initial Value Problem for a Third-Order Euler–Cauchy Equation
Solve the following initial value problem on any open interval I on the positive x-axis containing x ⫽ 1.
x 3y t ⫺ 3x 2y s ⫹ 6xy r ⫺ 6y ⫽ 0,
Solution.
y(1) ⫽ 2,
y r (1) ⫽ 1,
y s (1) ⫽ ⫺4.
Step 1. General solution. As in Sec. 2.5 we try y ⫽ x m. By differentiation and substitution,
m(m ⫺ 1)(m ⫺ 2)x m ⫺ 3m(m ⫺ 1)x m ⫹ 6mx m ⫺ 6x m ⫽ 0.
Dropping x m and ordering gives m 3 ⫺ 6m 2 ⫹ 11m ⫺ 6 ⫽ 0. If we can guess the root m ⫽ 1. We can divide
by m ⫺ 1 and find the other roots 2 and 3, thus obtaining the solutions x, x 2, x 3, which are linearly independent
on I (see Example 2). [In general one shall need a root-finding method, such as Newton’s (Sec. 19.2), also
available in a CAS (Computer Algebra System).] Hence a general solution is
y ⫽ c1x ⫹ c2 x 2 ⫹ c3 x 3
valid on any interval I, even when it includes x ⫽ 0 where the coefficients of the ODE divided by x 3 (to have
the standard form) are not continuous.
Step 2. Particular solution. The derivatives are y r ⫽ c1 ⫹ 2c2 x ⫹ 3c3 x 2 and y s ⫽ 2c2 ⫹ 6c3 x. From this, and
y and the initial conditions, we get by setting x ⫽ 1
(a) y(1) ⫽ c1 ⫹ c2 ⫹ c3 ⫽
2
(b) y r (1) ⫽ c1 ⫹ 2c2 ⫹ 3c3 ⫽
1
(c) y s (1) ⫽
2c2 ⫹ 6c3 ⫽ ⫺4.
This is solved by Cramer’s rule (Sec. 7.6), or by elimination, which is simple, as follows. (b) ⫺ (a) gives
(d) c2 ⫹ 2c3 ⫽ ⫺1. Then (c) ⫺ 2(d) gives c3 ⫽ ⫺1. Then (c) gives c2 ⫽ 1. Finally c1 ⫽ 2 from (a).
䊏
Answer: y ⫽ 2x ⫹ x 2 ⫺ x 3.
Linear Independence of Solutions. Wronskian
Linear independence of solutions is crucial for obtaining general solutions. Although it can
often be seen by inspection, it would be good to have a criterion for it. Now Theorem 2
in Sec. 2.6 extends from order n ⫽ 2 to any n. This extended criterion uses the Wronskian
W of n solutions y1, Á , yn defined as the nth-order determinant
(6)
W(y1, Á , yn) ⫽ 5
y1
y2
Á
yn
y1r
y2r
Á
ynr
#
#
Á
#
Á
y (nⴚ1)
n
y (nⴚ1)
y (nⴚ1)
1
2
5.
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SEC. 3.1 Homogeneous Linear ODEs
109
Note that W depends on x since y1, Á , yn do. The criterion states that these solutions
form a basis if and only if W is not zero; more precisely:
THEOREM 3
Linear Dependence and Independence of Solutions
Let the ODE (2) have continuous coefficients p0(x), Á , pnⴚ1(x) on an open interval
I. Then n solutions y1, Á , yn of (2) on I are linearly dependent on I if and only if their
Wronskian is zero for some x ⫽ x 0 in I. Furthermore, if W is zero for x ⫽ x 0, then W
is identically zero on I. Hence if there is an x 1 in I at which W is not zero, then y1, Á , yn
are linearly independent on I, so that they form a basis of solutions of (2) on I.
PROOF
(a) Let y1, Á , yn be linearly dependent solutions of (2) on I. Then, by definition, there
are constants k 1, Á , k n not all zero, such that for all x in I,
k 1 y1 ⫹ Á ⫹ k n yn ⫽ 0.
(7)
By n ⫺ 1 differentiations of (7) we obtain for all x in I
k 1 y1r ⫹ Á ⫹ k n ynr
⫽0
.
.
.
(8)
k 1y (nⴚ1)
⫹ Á ⫹ k ny (nⴚ1)
⫽ 0.
1
n
(7), (8) is a homogeneous linear system of algebraic equations with a nontrivial solution
k 1, Á , k n. Hence its coefficient determinant must be zero for every x on I, by Cramer’s
theorem (Sec. 7.7). But that determinant is the Wronskian W, as we see from (6). Hence
W is zero for every x on I.
(b) Conversely, if W is zero at an x 0 in I, then the system (7), (8) with x ⫽ x 0 has a
solution k 1*, Á , k n*, not all zero, by the same theorem. With these constants we define
the solution y* ⫽ k 1*y1 ⫹ Á ⫹ k n* yn of (2) on I. By (7), (8) this solution satisfies the
initial conditions y*(x 0) ⫽ 0, Á , y*(nⴚ1)(x 0) ⫽ 0. But another solution satisfying the
same conditions is y ⬅ 0. Hence y* ⬅ y by Theorem 2, which applies since the coefficients
of (2) are continuous. Together, y* ⫽ k 1*y1 ⫹ Á ⫹ k n* yn ⬅ 0 on I. This means linear
dependence of y1, Á , yn on I.
(c) If W is zero at an x 0 in I, we have linear dependence by (b) and then W ⬅ 0 by (a).
Hence if W is not zero at an x 1 in I, the solutions y1, Á , yn must be linearly independent
on I.
䊏
EXAMPLE 5
Basis, Wronskian
We can now prove that in Example 3 we do have a basis. In evaluating W, pull out the exponential functions
columnwise. In the result, subtract Column 1 from Columns 2, 3, 4 (without changing Column 1). Then expand by
Row 1. In the resulting third-order determinant, subtract Column 1 from Column 2 and expand the result by Row 2:
W⫽6
eⴚ2x
eⴚx
ex
e2x
1
1
1
1
⫺2eⴚ2x
⫺eⴚx
ex
2e2x
⫺2
⫺1
1
2
4eⴚ2x
eⴚx
ex
4e2x
4
1
1
4
⫺8eⴚ2x
⫺eⴚx
ex
8e2x
⫺8
⫺1
1
8
6⫽6
1
3
6 ⫽ 3 ⫺3
⫺3
7
9
4
0 3 ⫽ 72.
16
䊏
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CHAP. 3 Higher Order Linear ODEs
A General Solution of (2) Includes All Solutions
Let us first show that general solutions always exist. Indeed, Theorem 3 in Sec. 2.6 extends
as follows.
THEOREM 4
Existence of a General Solution
If the coefficients p0(x), Á , pnⴚ1(x) of (2) are continuous on some open interval I,
then (2) has a general solution on I.
PROOF
We choose any fixed x 0 in I. By Theorem 2 the ODE (2) has n solutions y1, Á , yn, where
yj satisfies initial conditions (5) with K jⴚ1 ⫽ 1 and all other K’s equal to zero. Their
Wronskian at x 0 equals 1. For instance, when n ⫽ 3, then y1(x 0) ⫽ 1, y2r (x 0) ⫽ 1,
y3s (x 0) ⫽ 1, and the other initial values are zero. Thus, as claimed,
y1(x 0)
y2(x 0)
y3(x 0)
1
0
0
W( y1(x 0), y2(x 0), y3(x 0)) ⫽ 4 y1r (x 0)
y2r (x 0)
y3r (x 0) 4 ⫽ 4 0
1
0 4 ⫽ 1.
y1s (x 0)
y2s (x 0)
y3s (x 0)
0
1
0
Hence for any n those solutions y1, Á , yn are linearly independent on I, by Theorem 3.
They form a basis on I, and y ⫽ c1 y1 ⫹ Á ⫹ cn yn is a general solution of (2) on I. 䊏
We can now prove the basic property that, from a general solution of (2), every solution
of (2) can be obtained by choosing suitable values of the arbitrary constants. Hence an
nth-order linear ODE has no singular solutions, that is, solutions that cannot be obtained
from a general solution.
THEOREM 5
General Solution Includes All Solutions
If the ODE (2) has continuous coefficients p0(x), Á , pnⴚ1(x) on some open interval
I, then every solution y ⫽ Y(x) of (2) on I is of the form
(9)
Y(x) ⫽ C1 y1(x) ⫹ Á ⫹ Cn yn(x)
where y1, Á , yn is a basis of solutions of (2) on I and C1, Á , Cn are suitable constants.
PROOF
Let Y be a given solution and y ⫽ c1 y1 ⫹ Á ⫹ cn yn a general solution of (2) on I. We
choose any fixed x 0 in I and show that we can find constants c1, Á , cn for which y and
its first n ⫺ 1 derivatives agree with Y and its corresponding derivatives at x 0. That is,
we should have at x ⫽ x 0
(10)
c1 y1 ⫹ Á ⫹
cn yn
⫽Y
c1 y1r ⫹ Á ⫹
cn ynr
⫽ Yr
.
.
.
c1 y (nⴚ1)
⫹ Á ⫹ cn y (nⴚ1)
⫽ Y (nⴚ1).
1
n
But this is a linear system of equations in the unknowns c1, Á , cn. Its coefficient
determinant is the Wronskian W of y1, Á , yn at x 0. Since y1, Á , yn form a basis, they
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SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients
111
are linearly independent, so that W is not zero by Theorem 3. Hence (10) has a unique
solution c1 ⫽ C1, Á , cn ⫽ Cn (by Cramer’s theorem in Sec. 7.7). With these values we
obtain the particular solution
y*(x) ⫽ C1 y1(x) ⫹ Á ⫹ Cn yn(x)
on I. Equation (10) shows that y* and its first n ⫺ 1 derivatives agree at x 0 with Y and
its corresponding derivatives. That is, y* and Y satisfy, at x 0, the same initial conditions.
The uniqueness theorem (Theorem 2) now implies that y* ⬅ Y on I. This proves the
theorem.
䊏
This completes our theory of the homogeneous linear ODE (2). Note that for n ⫽ 2 it is
identical with that in Sec. 2.6. This had to be expected.
PROBLEM SET 3.1
1–6
BASES: TYPICAL EXAMPLES
To get a feel for higher order ODEs, show that the given
functions are solutions and form a basis on any interval.
Use Wronskians. In Prob. 6, x ⬎ 0,
1. 1, x, x 2, x 3, y iv ⫽ 0
2. ex, eⴚx, e2x, y t ⫺ 2y s ⫺ y r ⫹ 2y ⫽ 0
3. cos x, sin x, x cos x, x sin x, y iv ⫹ 2y s ⫹ y ⫽ 0
4. eⴚ4x, xeⴚ4x, x 2eⴚ4x, y t ⫹ 12y s ⫹ 48y r ⫹ 64y ⫽ 0
5. 1, eⴚx cos 2x, eⴚx sin 2x, y t ⫹ 2y s ⫹ 5y r ⫽ 0
6. 1, x 2, x 4, x 2y t ⫺ 3xy s ⫹ 3y r ⫽ 0
7. TEAM PROJECT. General Properties of Solutions
of Linear ODEs. These properties are important in
obtaining new solutions from given ones. Therefore
extend Team Project 38 in Sec. 2.2 to nth-order ODEs.
Explore statements on sums and multiples of solutions
of (1) and (2) systematically and with proofs.
Recognize clearly that no new ideas are needed in this
extension from n ⫽ 2 to general n.
8–15
LINEAR INDEPENDENCE
Are the given functions linearly independent or dependent
on the half-axis x ⱖ 0? Give reason.
8. x 2, 1>x 2, 0
9. tan x, cot x, 1
10. e2x, xe2x, x 2e2x
11. ex cos x, ex sin x, ex
12. sin2 x, cos2 x, cos 2x
13. sin x, cos x, sin 2x
2
2
14. cos x, sin x, 2p
15. cosh 2x, sinh 2x, e2x
16. TEAM PROJECT. Linear Independence and
Dependence. (a) Investigate the given question about
a set S of functions on an interval I. Give an example.
Prove your answer.
(1) If S contains the zero function, can S be linearly
independent?
(2) If S is linearly independent on a subinterval J of I,
is it linearly independent on I?
(3) If S is linearly dependent on a subinterval J of I,
is it linearly dependent on I?
(4) If S is linearly independent on I, is it linearly
independent on a subinterval J?
(5) If S is linearly dependent on I, is it linearly
independent on a subinterval J?
(6) If S is linearly dependent on I, and if T contains S,
is T linearly dependent on I?
(b) In what cases can you use the Wronskian for
testing linear independence? By what other means can
you perform such a test?
3.2 Homogeneous Linear ODEs
with Constant Coefficients
We proceed along the lines of Sec. 2.2, and generalize the results from n ⫽ 2 to arbitrary n.
We want to solve an nth-order homogeneous linear ODE with constant coefficients,
written as
(1)
y (n) ⫹ anⴚ1 y (nⴚ1) ⫹ Á ⫹ a1 y r ⫹ a0y ⫽ 0
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CHAP. 3 Higher Order Linear ODEs
where y (n) ⫽ d ny>dx n, etc. As in Sec. 2.2, we substitute y ⫽ elx to obtain the characteristic
equation
(2)
l(n) ⫹ anⴚ1l(nⴚ1) ⫹ Á ⫹ a1l ⫹ a0y ⫽ 0
of (1). If l is a root of (2), then y ⫽ elx is a solution of (1). To find these roots, you may
need a numeric method, such as Newton’s in Sec. 19.2, also available on the usual CASs.
For general n there are more cases than for n ⫽ 2. We can have distinct real roots, simple
complex roots, multiple roots, and multiple complex roots, respectively. This will be shown
next and illustrated by examples.
Distinct Real Roots
If all the n roots l1, Á , ln of (2) are real and different, then the n solutions
(3)
y1 ⫽ el1x,
yn ⫽ elnx.
Á,
constitute a basis for all x. The corresponding general solution of (1) is
(4)
y ⫽ c1el1x ⫹ Á ⫹ cnelnx.
Indeed, the solutions in (3) are linearly independent, as we shall see after the example.
EXAMPLE 1
Distinct Real Roots
Solve the ODE y t ⫺ 2y s ⫺ y r ⫹ 2y ⫽ 0.
The characteristic equation is l3 ⫺ 2l2 ⫺ l ⫹ 2 ⫽ 0. It has the roots ⫺1, 1, 2; if you find one
of them by inspection, you can obtain the other two roots by solving a quadratic equation (explain!). The
corresponding general solution (4) is y ⫽ c1eⴚx ⫹ c2ex ⫹ c3e2x.
䊏
Solution.
Linear Independence of (3). Students familiar with nth-order determinants may verify
that, by pulling out all exponential functions from the columns and denoting their product
by E ⫽ exp [l1 ⫹ Á ⫹ ln)x], the Wronskian of the solutions in (3) becomes
(5)
el1x
el2x
Á
elnx
l1el1x
l2el2x
Á
lnelnx
W ⫽ 7 l21el1x
l22el2x
Á
l2nelnx 7
#
#
Á
#
lnⴚ1
el1x
1
lnⴚ1
el2x
2
Á
lnⴚ1
elnx
n
1
1
Á
1
l1
l2
Á
ln
⫽ E 7 l21
l22
Á
l2n 7 .
#
#
Á
#
lnⴚ1
1
lnⴚ1
2
Á
lnⴚ1
n
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SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients
113
The exponential function E is never zero. Hence W ⫽ 0 if and only if the determinant on
the right is zero. This is a so-called Vandermonde or Cauchy determinant.1 It can be
shown that it equals
(⫺1)n(nⴚ1)>2V
(6)
where V is the product of all factors lj ⫺ lk with j ⬍ k (⬉ n); for instance, when n ⫽ 3
we get ⫺V ⫽ ⫺(l1 ⫺ l2)(l1 ⫺ l3)(l2 ⫺ l3). This shows that the Wronskian is not zero
if and only if all the n roots of (2) are different and thus gives the following.
THEOREM 1
Basis
Solutions y1 ⫽ el1x, Á , yn ⫽ elnx of (1) (with any real or complex lj’s) form a
basis of solutions of (1) on any open interval if and only if all n roots of (2) are
different.
Actually, Theorem 1 is an important special case of our more general result obtained
from (5) and (6):
THEOREM 2
Linear Independence
Any number of solutions of (1) of the form elx are linearly independent on an open
interval I if and only if the corresponding l are all different.
Simple Complex Roots
If complex roots occur, they must occur in conjugate pairs since the coefficients of (1)
are real. Thus, if l ⫽ g ⫹ iv is a simple root of (2), so is the conjugate l ⫽ g ⫺ iv, and
two corresponding linearly independent solutions are (as in Sec. 2.2, except for notation)
y1 ⫽ egx cos vx,
EXAMPLE 2
y2 ⫽ egx sin vx.
Simple Complex Roots. Initial Value Problem
Solve the initial value problem
y t ⫺ y s ⫹ 100y r ⫺ 100y ⫽ 0,
y(0) ⫽ 4,
y r (0) ⫽ 11,
y s (0) ⫽ ⫺299.
The characteristic equation is l3 ⫺ l2 ⫹ 100l ⫺ 100 ⫽ 0. It has the root 1, as can perhaps be
seen by inspection. Then division by l ⫺ 1 shows that the other roots are ⫾10i. Hence a general solution and
its derivatives (obtained by differentiation) are
Solution.
y ⫽ c1ex ⫹ A cos 10x ⫹ B sin 10x,
y r ⫽ c1ex ⫺ 10A sin 10x ⫹ 10B cos 10x,
y s ⫽ c1ex ⫺ 100A cos 10x ⫺ 100B sin 10x.
1
ALEXANDRE THÉOPHILE VANDERMONDE (1735–1796), French mathematician, who worked on
solution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5.
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CHAP. 3 Higher Order Linear ODEs
From this and the initial conditions we obtain, by setting x ⫽ 0,
(a) c1 ⫹ A ⫽ 4,
(b) c1 ⫹ 10B ⫽ 11,
(c) c1 ⫺ 100A ⫽ ⫺299.
We solve this system for the unknowns A, B, c1. Equation (a) minus Equation (c) gives 101A ⫽ 303, A ⫽ 3.
Then c1 ⫽ 1 from (a) and B ⫽ 1 from (b). The solution is (Fig. 73)
y ⫽ ex ⫹ 3 cos 10x ⫹ sin 10x.
This gives the solution curve, which oscillates about ex (dashed in Fig. 73).
䊏
y
20
10
4
0
0
1
2
3
x
Fig. 73. Solution in Example 2
Multiple Real Roots
If a real double root occurs, say, l1 ⫽ l2, then y1 ⫽ y2 in (3), and we take y1 and xy1 as
corresponding linearly independent solutions. This is as in Sec. 2.2.
More generally, if l is a real root of order m, then m corresponding linearly independent
solutions are
(7)
elx,
xelx,
x 2elx,
Á , x mⴚ1elx.
We derive these solutions after the next example and indicate how to prove their linear
independence.
EXAMPLE 3
Real Double and Triple Roots
Solve the ODE y v ⫺ 3y iv ⫹ 3y t ⫺ y s ⫽ 0.
The characteristic equation l5 ⫺ 3l4 ⫹ 3l3 ⫺ l2 ⫽ 0 has the roots l1 ⫽ l2 ⫽ 0, and l3 ⫽ l4 ⫽
l5 ⫽ 1, and the answer is
Solution.
(8)
y ⫽ c1 ⫹ c2 x ⫹ (c3 ⫹ c4 x ⫹ c5 x 2)ex.
Derivation of (7). We write the left side of (1) as
L[ y] ⫽ y (n) ⫹ anⴚ1 y (nⴚ1) ⫹ Á ⫹ a0y.
Let y ⫽ elx. Then by performing the differentiations we have
L[elx] ⫽ (ln ⫹ anⴚ1lnⴚ1 ⫹ Á ⫹ a0)elx.
䊏
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SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients
115
Now let l1 be a root of mth order of the polynomial on the right, where m ⬉ n. For m ⬍ n
let lmⴙ1, Á , ln be the other roots, all different from l1. Writing the polynomial in
product form, we then have
L[elx] ⫽ (l ⫺ l1)mh(l)elx
with h(l) ⫽ 1 if m ⫽ n, and h(l) ⫽ (l ⫺ lm⫹1) Á (l ⫺ ln) if m ⬍ n. Now comes the
key idea: We differentiate on both sides with respect to l,
(9)
0
0
L[elx] ⫽ m(l ⫺ l1)mⴚ1h(l)elx ⫹ (l ⫺ l1)m
[h(l)elx].
0l
0l
The differentiations with respect to x and l are independent and the resulting derivatives
are continuous, so that we can interchange their order on the left:
(10)
0
0 lx
L[elx] ⫽ L c
e d ⫽ L[xelx].
0l
0l
The right side of (9) is zero for l ⫽ l1 because of the factors l ⫺ l1 (and m ⭌ 2 since
we have a multiple root!). Hence L[xel1x] ⫽ 0 by (9) and (10). This proves that xel1x is
a solution of (1).
We can repeat this step and produce x 2el1x, Á , x mⴚ1el1x by another m ⫺ 2 such
differentiations with respect to l. Going one step further would no longer give zero on the
right because the lowest power of l ⫺ l1 would then be (l ⫺ l1)0, multiplied by m!h(l)
and h(l1) ⫽ 0 because h(l) has no factors l ⫺ l1; so we get precisely the solutions in (7).
We finally show that the solutions (7) are linearly independent. For a specific n this
can be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrary
m we can pull out the exponential functions from the Wronskian. This gives (elx)m ⫽ elmx
times a determinant which by “row operations” can be reduced to the Wronskian of 1,
x, Á , x mⴚ1. The latter is constant and different from zero (equal to 1!2! Á (m ⫺ 1)!).
These functions are solutions of the ODE y (m) ⫽ 0, so that linear independence follows
from Theroem 3 in Sec. 3.1.
Multiple Complex Roots
In this case, real solutions are obtained as for complex simple roots above. Consequently,
if l ⫽ g ⫹ iv is a complex double root, so is the conjugate l ⫽ g ⫺ iv. Corresponding
linearly independent solutions are
(11)
egx cos vx,
egx sin vx,
xegx cos vx,
xegx sin vx.
The first two of these result from elx and elx as before, and the second two from xelx
and xelx in the same fashion. Obviously, the corresponding general solution is
(12)
y ⫽ egx[(A1 ⫹ A2x) cos vx ⫹ (B1 ⫹ B2x) sin vx].
For complex triple roots (which hardly ever occur in applications), one would obtain
two more solutions x 2egx cos vx, x 2egx sin vx, and so on.
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CHAP. 3 Higher Order Linear ODEs
PROBLEM SET 3.2
1–6
GENERAL SOLUTION
Solve the given ODE. Show the details of your work.
1. y t ⫹ 25y r ⫽ 0
2. y iv ⫹ 2y s ⫹ y ⫽ 0
3. y iv ⫹ 4y s ⫽ 0
4. (D 3 ⫺ D 2 ⫺ D ⫹ I ) y ⫽ 0
5. (D 4 ⫹ 10D 2 ⫹ 9I ) y ⫽ 0
6. (D 5 ⫹ 8D 3 ⫹ 16D) y ⫽ 0
7–13
INITIAL VALUE PROBLEM
Solve the IVP by a CAS, giving a general solution and the
particular solution and its graph.
7. y t ⫹ 3.2y s ⫹ 4.81y r ⫽ 0, y(0) ⫽ 3.4, y r(0) ⫽ ⫺4.6,
y s (0) ⫽ 9.91
8. y t ⫹ 7.5y s ⫹ 14.25y r ⫺ 9.125y ⫽ 0, y(0) ⫽ 10.05,
y r (0) ⫽ ⫺54.975, y s (0) ⫽ 257.5125
9. 4y t ⫹ 8y s ⫹ 41y r ⫹ 37y ⫽ 0, y(0) ⫽ 9,
y r (0) ⫽ ⫺6.5, y s (0) ⫽ ⫺39.75
10. y iv ⫹ 4y ⫽ 0, y(0) ⫽ 12, y r (0) ⫽ ⫺ 32, y s (0) ⫽ 52,
y t (0) ⫽ ⫺72
11. y iv ⫺ 9y s ⫺ 400y ⫽ 0, y(0) ⫽ 0, y r (0) ⫽ 0,
y s (0) ⫽ 41, y t (0) ⫽ 0
12. y v ⫺ 5y t ⫹ 4y r ⫽ 0, y(0) ⫽ 3, y r (0) ⫽ ⫺5,
y s (0) ⫽ 11, y t (0) ⫽ ⫺23, y iv(0) ⫽ 47
13. y iv ⫹ 0.45y t ⫺ 0.165y s ⫹ 0.0045y r ⫺ 0.00175y ⫽ 0,
y(0) ⫽ 17.4, y r (0) ⫽ ⫺2.82, y s (0) ⫽ 2.0485,
y t (0) ⫽ ⫺1.458675
14. PROJECT. Reduction of Order. This is of practical
interest since a single solution of an ODE can often be
guessed. For second order, see Example 7 in Sec. 2.1.
(a) How could you reduce the order of a linear
constant-coefficient ODE if a solution is known?
(b) Extend the method to a variable-coefficient ODE
y t ⫹ p2(x)y s ⫹ p1(x)y r ⫹ p0(x)y ⫽ 0.
Assuming a solution y1 to be known, show that another
solution is y2(x) ⫽ u(x)y1(x) with u(x) ⫽ 兰 z(x) dx and
z obtained by solving
y1z s ⫹ (3y1r ⫹ p2 y1)z r ⫹ (3y1s ⫹ 2p2 y1r ⫹ p1 y1)z ⫽ 0.
(c) Reduce
x 3y t ⫺ 3x 2y s ⫹ (6 ⫺ x 2)xy r ⫺ (6 ⫺ x 2)y ⫽ 0,
using y1 ⫽ x (perhaps obtainable by inspection).
15. CAS EXPERIMENT. Reduction of Order. Starting
with a basis, find third-order linear ODEs with variable
coefficients for which the reduction to second order
turns out to be relatively simple.
3.3 Nonhomogeneous Linear ODEs
We now turn from homogeneous to nonhomogeneous linear ODEs of nth order. We write
them in standard form
(1)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ r (x)
with y (n) ⫽ d ny>dx n as the first term, and r (x) [ 0. As for second-order ODEs, a general
solution of (1) on an open interval I of the x-axis is of the form
(2)
y(x) ⫽ yh(x) ⫹ yp(x).
Here yh(x) ⫽ c1 y1(x) ⫹ Á ⫹ cn yn(x) is a general solution of the corresponding
homogeneous ODE
(3)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ 0
on I. Also, yp is any solution of (1) on I containing no arbitrary constants. If (1) has
continuous coefficients and a continuous r (x) on I, then a general solution of (1) exists
and includes all solutions. Thus (1) has no singular solutions.
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SEC. 3.3 Nonhomogeneous Linear ODEs
117
An initial value problem for (1) consists of (1) and n initial conditions
y(x 0) ⫽ K 0,
(4)
y r (x 0) ⫽ K 1,
y (nⴚ1)(x 0) ⫽ K nⴚ1
Á,
with x 0 in I. Under those continuity assumptions it has a unique solution. The ideas of
proof are the same as those for n ⫽ 2 in Sec. 2.7.
Method of Undetermined Coefficients
Equation (2) shows that for solving (1) we have to determine a particular solution of (1).
For a constant-coefficient equation
y (n) ⫹ anⴚ1 y (nⴚ1) ⫹ Á ⫹ a1 y r ⫹ a0y ⫽ r (x)
(5)
(a0, Á , anⴚ1 constant) and special r (x) as in Sec. 2.7, such a yp(x) can be determined by
the method of undetermined coefficients, as in Sec. 2.7, using the following rules.
(A) Basic Rule as in Sec. 2.7.
(B) Modification Rule. If a term in your choice for yp(x) is a solution of the
homogeneous equation (3), then multiply this term by x k, where k is the smallest
positive integer such that this term times x k is not a solution of (3).
(C) Sum Rule as in Sec. 2.7.
The practical application of the method is the same as that in Sec. 2.7. It suffices to
illustrate the typical steps of solving an initial value problem and, in particular, the new
Modification Rule, which includes the old Modification Rule as a particular case (with
k ⫽ 1 or 2). We shall see that the technicalities are the same as for n ⫽ 2, except perhaps
for the more involved determination of the constants.
EXAMPLE 1
Initial Value Problem. Modification Rule
Solve the initial value problem
(6)
y t ⫹ 3y s ⫹ 3y r ⫹ y ⫽ 30eⴚx,
y(0) ⫽ 3,
y r (0) ⫽ ⫺3,
y s (0) ⫽ ⫺47.
Step 1. The characteristic equation is l3 ⫹ 3l2 ⫹ 3l ⫹ 1 ⫽ (l ⫹ 1)3 ⫽ 0. It has the triple root
l ⫽ ⫺1. Hence a general solution of the homogeneous ODE is
Solution.
yh ⫽ c1eⴚx ⫹ c2 xeⴚx ⫹ c3 x 2eⴚx
⫽ (c1 ⫹ c2 x ⫹ c3 x 2)eⴚx.
Step 2. If we try yp ⫽ Ceⴚx, we get ⫺C ⫹ 3C ⫺ 3C ⫹ C ⫽ 30, which has no solution. Try Cxeⴚx and
Cx 2eⴚx. The Modification Rule calls for
yp ⫽ Cx 3eⴚx.
Then
ypr ⫽ C(3x 2 ⫺ x 3)eⴚx,
yps ⫽ C(6x ⫺ 6x 2 ⫹ x 3)eⴚx,
ypt ⫽ C(6 ⫺ 18x ⫹ 9x 2 ⫺ x 3)eⴚx.
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CHAP. 3 Higher Order Linear ODEs
Substitution of these expressions into (6) and omission of the common factor eⴚx gives
C(6 ⫺ 18x ⫹ 9x 2 ⫺ x 3) ⫹ 3C(6x ⫺ 6x 2 ⫹ x 3) ⫹ 3C(3x 2 ⫺ x 3) ⫹ Cx 3 ⫽ 30.
The linear, quadratic, and cubic terms drop out, and 6C ⫽ 30. Hence C ⫽ 5. This gives yp ⫽ 5x 3eⴚx.
Step 3. We now write down y ⫽ yh ⫹ yp, the general solution of the given ODE. From it we find c1 by the
first initial condition. We insert the value, differentiate, and determine c2 from the second initial condition, insert
the value, and finally determine c3 from y s (0) and the third initial condition:
y ⫽ yh ⫹ yp ⫽ (c1 ⫹ c2x ⫹ c3x 2)eⴚx ⫹ 5x 3eⴚx,
y(0) ⫽ c1 ⫽ 3
y r ⫽ [⫺3 ⫹ c2 ⫹ (⫺c2 ⫹ 2c3)x ⫹ (15 ⫺ c3)x 2 ⫺ 5x 3]eⴚx,
y r (0) ⫽ ⫺3 ⫹ c2 ⫽ ⫺3,
c2 ⫽ 0
y s ⫽ [3 ⫹ 2c3 ⫹ (30 ⫺ 4c3)x ⫹ (⫺30 ⫹ c3)x 2 ⫹ 5x 3]eⴚx,
y s (0) ⫽ 3 ⫹ 2c3 ⫽ ⫺47,
c3 ⫽ ⫺25.
Hence the answer to our problem is (Fig. 73)
y ⫽ (3 ⫺ 25x 2)eⴚx ⫹ 5x 3eⴚx.
The curve of y begins at (0, 3) with a negative slope, as expected from the initial values, and approaches zero
as x : ⬁. The dashed curve in Fig. 74 is yp.
䊏
y
5
0
5
10
x
–5
Fig. 74. y and yp (dashed) in Example 1
Method of Variation of Parameters
The method of variation of parameters (see Sec. 2.10) also extends to arbitrary order n.
It gives a particular solution yp for the nonhomogeneous equation (1) (in standard form
with y (n) as the first term!) by the formula
n
yp(x) ⫽ a yk(x)
k⫽1
(7)
⫽ y1(x)
冮 W(x) r (x) dx
Wk(x)
冮 W(x) r (x) dx ⫹ Á ⫹ y (x) 冮 W(x) r (x) dx
W1(x)
Wn(x)
n
on an open interval I on which the coefficients of (1) and r (x) are continuous. In (7) the
functions y1, Á , yn form a basis of the homogeneous ODE (3), with Wronskian W, and
Wj ( j ⫽ 1, Á , n) is obtained from W by replacing the jth column of W by the column
[0 0 Á 0 1]T. Thus, when n ⫽ 2, this becomes identical with (2) in Sec. 2.10,
W⫽ `
y1
y2
y1r
y2r
`,
W1 ⫽ `
0
y2
1
y2r
` ⫽ ⫺y2,
W2 ⫽ `
y1
0
y1r
1
` ⫽ y1.
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SEC. 3.3 Nonhomogeneous Linear ODEs
119
The proof of (7) uses an extension of the idea of the proof of (2) in Sec. 2.10 and can
be found in Ref [A11] listed in App. 1.
EXAMPLE 2
Variation of Parameters. Nonhomogeneous Euler–Cauchy Equation
Solve the nonhomogeneous Euler–Cauchy equation
x 3y t ⫺ 3x 2y s ⫹ 6xy r ⫺ 6y ⫽ x 4 ln x
(x ⬎ 0).
Step 1. General solution of the homogeneous ODE. Substitution of y ⫽ x m and the derivatives
into the homogeneous ODE and deletion of the factor x m gives
Solution.
m(m ⫺ 1)(m ⫺ 2) ⫺ 3m(m ⫺ 1) ⫹ 6m ⫺ 6 ⫽ 0.
The roots are 1, 2, 3 and give as a basis
y1 ⫽ x,
y2 ⫽ x 2,
y3 ⫽ x 3.
Hence the corresponding general solution of the homogeneous ODE is
yh ⫽ c1x ⫹ c2x 2 ⫹ c3x 3.
Step 2. Determinants needed in (7). These are
x
x2
x3
W⫽31
2x
3x 2 3 ⫽ 2x 3
0
2
6x
0
x2
x3
W1 ⫽ 4 0
2x
3x 2 4 ⫽ x 4
1
2
6x
x
0
x3
W2 ⫽ 4 1
0
3x 2 4 ⫽ ⫺2x 3
0
1
6x
x
x2
0
W3 ⫽ 4 1
2x
0 4 ⫽ x 2.
0
2
1
Step 3. Integration. In (7) we also need the right side r (x) of our ODE in standard form, obtained by division
of the given equation by the coefficient x 3 of y t ; thus, r (x) ⫽ (x 4 ln x)>x 3 ⫽ x ln x. In (7) we have the simple
quotients W1>W ⫽ x>2, W2>W ⫽ ⫺1, W3>W ⫽ 1>(2x). Hence (7) becomes
yp ⫽ x
⫽
冮 2 x ln x dx ⫺ x 冮 x ln x dx ⫹ x 冮 2x x ln x dx
x
2
3
1
x x3
x3
x2
x2
x3
a ln x ⫺ b ⫺ x 2 a ln x ⫺ b ⫹
(x ln x ⫺ x).
2 3
9
2
4
2
Simplification gives yp ⫽ 16 x 4 (ln x ⫺ 11
6 ). Hence the answer is
y ⫽ yh ⫹ yp ⫽ c1x ⫹ c2 x 2 ⫹ c3 x 3 ⫹ 16 x 4 (ln x ⫺ 11
6 ).
Figure 75 shows yp. Can you explain the shape of this curve? Its behavior near x ⫽ 0? The occurrence of a minimum?
䊏
Its rapid increase? Why would the method of undetermined coefficients not have given the solution?
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CHAP. 3 Higher Order Linear ODEs
y
30
20
10
0
x
10
5
–10
–20
Fig. 75. Particular solution yp of the nonhomogeneous
Euler–Cauchy equation in Example 2
Application: Elastic Beams
Whereas second-order ODEs have various applications, of which we have discussed some
of the more important ones, higher order ODEs have much fewer engineering applications.
An important fourth-order ODE governs the bending of elastic beams, such as wooden or
iron girders in a building or a bridge.
A related application of vibration of beams does not fit in here since it leads to PDEs
and will therefore be discussed in Sec. 12.3.
EXAMPLE 3
Bending of an Elastic Beam under a Load
We consider a beam B of length L and constant (e.g., rectangular) cross section and homogeneous elastic
material (e.g., steel); see Fig. 76. We assume that under its own weight the beam is bent so little that it is
practically straight. If we apply a load to B in a vertical plane through the axis of symmetry (the x-axis in
Fig. 76), B is bent. Its axis is curved into the so-called elastic curve C (or deflection curve). It is shown in
elasticity theory that the bending moment M(x) is proportional to the curvature k(x) of C. We assume the bending
to be small, so that the deflection y(x) and its derivative y r (x) (determining the tangent direction of C) are small.
Then, by calculus, k ⫽ y s >(1 ⫹ y r 2)3>2 ⬇ y s . Hence
M(x) ⫽ EIy s (x).
EI is the constant of proportionality. E is Young’s modulus of elasticity of the material of the beam. I is the
moment of inertia of the cross section about the (horizontal) z-axis in Fig. 76.
Elasticity theory shows further that M s (x) ⫽ f (x), where f (x) is the load per unit length. Together,
EIy iv ⫽ f (x).
(8)
x
L
y
z
Undeformed beam
x
y
z
Deformed beam
under uniform load
(simply supported)
Fig. 76. Elastic beam
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SEC. 3.3 Nonhomogeneous Linear ODEs
121
In applications the most important supports and corresponding boundary conditions are as follows and shown
in Fig. 77.
(A) Simply supported
y ⫽ y s ⫽ 0 at x ⫽ 0 and L
(B) Clamped at both ends
y ⫽ y r ⫽ 0 at x ⫽ 0 and L
(C) Clamped at x ⫽ 0, free at x ⫽ L
y(0) ⫽ y r (0) ⫽ 0, y s (L) ⫽ y t (L) ⫽ 0.
The boundary condition y ⫽ 0 means no displacement at that point, y r ⫽ 0 means a horizontal tangent, y s ⫽ 0
means no bending moment, and y t ⫽ 0 means no shear force.
Let us apply this to the uniformly loaded simply supported beam in Fig. 76. The load is f (x) ⬅ f0 ⫽ const.
Then (8) is
f0
(9)
y iv ⫽ k,
k⫽ .
EI
This can be solved simply by calculus. Two integrations give
k 2
x ⫹ c1x ⫹ c2.
2
ys ⫽
y s (0) ⫽ 0 gives c2 ⫽ 0. Then y s (L) ⫽ L (12 kL ⫹ c1) ⫽ 0, c1 ⫽ ⫺kL>2 (since L ⫽ 0). Hence
ys ⫽
k 2
(x ⫺ Lx).
2
Integrating this twice, we obtain
y⫽
k 1 4 L 3
a x ⫺ x ⫹ c3 x ⫹ c4 b
2 12
6
with c4 ⫽ 0 from y(0) ⫽ 0. Then
y(L) ⫽
kL L3
L3
a ⫺
⫹ c3 b ⫽ 0,
2 12
6
c3 ⫽
L3
.
12
Inserting the expression for k, we obtain as our solution
y⫽
f0
24EI
(x 4 ⫺ 2L x 3 ⫹ L3x).
Since the boundary conditions at both ends are the same, we expect the deflection y(x) to be “symmetric” with
respect to L>2, that is, y(x) ⫽ y(L ⫺ x). Verify this directly or set x ⫽ u ⫹ L>2 and show that y becomes an
even function of u,
y⫽
f0
24EI
au 2 ⫺
1 2
5
L b au 2 ⫺ L2 b .
4
4
From this we can see that the maximum deflection in the middle at u ⫽ 0 (x ⫽ L>2) is 5f0L4>(16 # 24EI). Recall
that the positive direction points downward.
䊏
x
(A) Simply supported
x=0
x=L
(B) Clamped at both
ends
x=0
x=0
x=L
x=L
(C) Clamped at the left
end, free at the
right end
Fig. 77. Supports of a beam
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CHAP. 3 Higher Order Linear ODEs
PROBLEM SET 3.3
1–7
GENERAL SOLUTION
Solve the following ODEs, showing the details of your
work.
1. y t ⫹ 3y s ⫹ 3y r ⫹ y ⫽ ex ⫺ x ⫺ 1
2. y t ⫹ 2y s ⫺ y r ⫺ 2y ⫽ 1 ⫺ 4x 3
3. (D 4 ⫹ 10D 2 ⫹ 9I ) y ⫽ 6.5 sinh 2x
4. (D 3 ⫹ 3D 2 ⫺ 5D ⫺ 39I )y ⫽ ⫺300 cos x
5. (x 3D 3 ⫹ x 2D 2 ⫺ 2xD ⫹ 2I )y ⫽ x ⴚ2
6. (D 3 ⫹ 4D)y ⫽ sin x
7. (D 3 ⫺ 9D 2 ⫹ 27D ⫺ 27I )y ⫽ 27 sin 3x
8–13
INITIAL VALUE PROBLEM
Solve the given IVP, showing the details of your work.
8. y iv ⫺ 5y s ⫹ 4y ⫽ 10eⴚ3x, y(0) ⫽ 1, y r (0) ⫽ 0,
y s (0) ⫽ 0, y t (0) ⫽ 0
9. y iv ⫹ 5y s ⫹ 4y ⫽ 90 sin 4x, y(0) ⫽ 1, y r (0) ⫽ 2,
y s (0) ⫽ ⫺1, y t (0) ⫽ ⫺32
10. x 3y t ⫹ xy r ⫺ y ⫽ x 2, y(1) ⫽ 1, y r (1) ⫽ 3,
y s (1) ⫽ 14
11. (D 3 ⫺ 2D 2 ⫺ 3D)y ⫽ 74eⴚ3x sin x, y(0) ⫽ ⫺1.4,
y r (0) ⫽ 3.2, y s (0) ⫽ ⫺5.2
12. (D 3 ⫺ 2D 2 ⫺ 9D ⫹ 18I )y ⫽ e2x, y(0) ⫽ 4.5,
y r (0) ⫽ 8.8, y s (0) ⫽ 17.2
13. (D 3 ⫺ 4D)y ⫽ 10 cos x ⫹ 5 sin x, y(0) ⫽ 3,
y r (0) ⫽ ⫺2, y s (0) ⫽ ⫺1
14. CAS EXPERIMENT. Undetermined Coefficients.
Since variation of parameters is generally complicated,
it seems worthwhile to try to extend the other method.
Find out experimentally for what ODEs this is possible
and for what not. Hint: Work backward, solving ODEs
with a CAS and then looking whether the solution
could be obtained by undetermined coefficients. For
example, consider
y t ⫺ 3y s ⫹ 3y r ⫺ y ⫽ x 1>2ex
and
x 3y t ⫹ x 2y s ⫺ 2xy r ⫹ 2y ⫽ x 3 ln x.
15. WRITING REPORT. Comparison of Methods. Write
a report on the method of undetermined coefficients and
the method of variation of parameters, discussing and
comparing the advantages and disadvantages of each
method. Illustrate your findings with typical examples.
Try to show that the method of undetermined coefficients,
say, for a third-order ODE with constant coefficients and
an exponential function on the right, can be derived from
the method of variation of parameters.
CHAPTER 3 REVIEW QUESTIONS AND PROBLEMS
1. What is the superposition or linearity principle? For
what nth-order ODEs does it hold?
2. List some other basic theorems that extend from
second-order to nth-order ODEs.
3. If you know a general solution of a homogeneous linear
ODE, what do you need to obtain from it a general
solution of a corresponding nonhomogeneous linear
ODE?
4. What form does an initial value problem for an nthorder linear ODE have?
5. What is the Wronskian? What is it used for?
6–15
GENERAL SOLUTION
Solve the given ODE. Show the details of your work.
6. y iv ⫺ 3y s ⫺ 4y ⫽ 0
7. y t ⫹ 4y s ⫹ 13y r ⫽ 0
8. y t ⫺ 4y s ⫺ y r ⫹ 4y ⫽ 30e2x
9. (D 4 ⫺ 16I )y ⫽ ⫺15 cosh x
10. x 2y t ⫹ 3xy s ⫺ 2y r ⫽ 0
11. y t ⫹ 4.5y s ⫹ 6.75y r ⫹ 3.375y ⫽ 0
12. (D 3 ⫺ D)y ⫽ sinh 0.8x
13. (D 3 ⫹ 6D 2 ⫹ 12D ⫹ 8I )y ⫽ 8x 2
14. (D 4 ⫺ 13D 2 ⫹ 36I )y ⫽ 12ex
15. 4x 3y t ⫹ 3xy r ⫺ 3y ⫽ 10
INITIAL VALUE PROBLEM
16–20
Solve the IVP. Show the details of your work.
16. (D 3 ⫺ D 2 ⫺ D ⫹ I )y ⫽ 0, y(0) ⫽ 0, Dy(0) ⫽ 1,
D 2y(0) ⫽ 0
17. y t ⫹ 5y s ⫹ 24y r ⫹ 20y ⫽ x, y(0) ⫽ 1.94,
y r (0) ⫽ ⫺3.95, y s ⫽ ⫺24
18. (D 4 ⫺ 26D 2 ⫹ 25I )y ⫽ 50(x ⫹ 1)2, y(0) ⫽ 12.16,
Dy(0) ⫽ ⫺6, D 2y(0) ⫽ 34, D 3y(0) ⫽ ⫺130
19. (D 3 ⫹ 9D 2 ⫹ 23D ⫹ 15I )y ⫽ 12exp(⫺4x),
y(0) ⫽ 9, Dy(0) ⫽ ⫺41, D 2y(0) ⫽ 189
20. (D 3 ⫹ 3D 2 ⫹ 3D ⫹ I )y ⫽ 8 sin x, y(0) ⫽ ⫺1,
y r (0) ⫽ ⫺3, y s (0) ⫽ 5
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Summary of Chapter 3
123
SUMMARY OF CHAPTER
3
Higher Order Linear ODEs
Compare with the similar Summary of Chap. 2 (the case n ⴝ 2).
Chapter 3 extends Chap. 2 from order n ⫽ 2 to arbitrary order n. An nth-order
linear ODE is an ODE that can be written
(1)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ r (x)
with y (n) ⫽ d ny>dx n as the first term; we again call this the standard form. Equation
(1) is called homogeneous if r (x) ⬅ 0 on a given open interval I considered,
nonhomogeneous if r (x) [ 0 on I. For the homogeneous ODE
(2)
y (n) ⫹ pnⴚ1(x)y (nⴚ1) ⫹ Á ⫹ p1(x)y r ⫹ p0(x)y ⫽ 0
the superposition principle (Sec. 3.1) holds, just as in the case n ⫽ 2. A basis or
fundamental system of solutions of (2) on I consists of n linearly independent
solutions y1, Á , yn of (2) on I. A general solution of (2) on I is a linear combination
of these,
(3)
y ⫽ c1 y1 ⫹ Á ⫹ cn yn
(c1, Á , cn arbitrary constants).
A general solution of the nonhomogeneous ODE (1) on I is of the form
y ⫽ yh ⫹ yp
(4)
(Sec. 3.3).
Here, yp is a particular solution of (1) and is obtained by two methods (undetermined
coefficients or variation of parameters) explained in Sec. 3.3.
An initial value problem for (1) or (2) consists of one of these ODEs and n
initial conditions (Secs. 3.1, 3.3)
(5)
y(x 0) ⫽ K 0,
y r (x 0) ⫽ K 1,
Á,
y (nⴚ1)(x 0) ⫽ K nⴚ1
with given x 0 in I and given K 0, Á , K nⴚ1. If p0, Á , pnⴚ1, r are continuous on I,
then general solutions of (1) and (2) on I exist, and initial value problems (1), (5)
or (2), (5) have a unique solution.